Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 22 Heights And Distances

Exercise 22A

Question 1. The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.
Answer:
Let the length of the shadow of the tree be x m.
∴ Height of the tree = \( \sqrt{3} \times m \)
If θ is the angle of elevation of the sun, then
\( \tan θ = \frac{\sqrt{3} \times x}{x} = \sqrt{3} = \tan 60° \)
∴ θ = 60°
In simple words: The sun makes a 60° angle with the ground. When the height is √3 times the shadow, we always get 60°.

[Diagram: This diagram shows a right triangle with the tree as the vertical side, shadow as the horizontal base, and the sun's ray making angle θ with the ground.]

📝 Teacher's Note: Use a stick in sunlight to show students how shadow changes with sun angle. When shadow is shorter, sun is higher up.

🎯 Exam Tip: Remember tan 60° = √3. Write this value clearly in your answer to get full marks.

Question 2. The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.
Answer:
Let the height of the tower be h m.
Given that angle of elevation is 60°
\( \tan 60° = \frac{h}{160} \)
\( \Rightarrow \sqrt{3} = \frac{h}{160} \)
∴ h = 160√3 = 277.12 m
So, height of the tower is 277.12 m.
In simple words: We stand 160 m away from a tower and look up at 60°. Using trigonometry, we find the tower is about 277 meters tall.

[Diagram: This diagram shows a right triangle with tower as vertical side, 160 m as horizontal distance, and 60° angle of elevation.]

📝 Teacher's Note: Show students how to use tan ratio. Height divided by base distance gives tan of the angle. This is the most common height-distance problem.

🎯 Exam Tip: Always write "Let height = h" first. Then write the tan formula. Substitute values carefully and show the final answer with units.

Question 3. A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68° with the ground. Find the height, upto which the ladder reaches.
Answer:
Let the height upto which the ladder reaches be h m.
Given that angle of elevation is 68°
\( \tan 68° = \frac{h}{2.4} \)
\( \Rightarrow 2.475 = \frac{h}{2.4} \)
∴ h = 2.475 × 2.4 = 5.94 m
So, the ladder reaches upto a height of 5.94 m.
In simple words: The ladder leans against a wall. Its bottom is 2.4 m away from the wall. The ladder makes 68° with ground, so it reaches 5.94 m up the wall.

[Diagram: This diagram shows a right triangle with the ladder as hypotenuse, wall height as vertical side, and 2.4 m as horizontal distance from wall.]

📝 Teacher's Note: Use a real ladder in class if possible. Show how the ladder, wall, and ground make a right triangle. Students understand this easily with a real example.

🎯 Exam Tip: Remember tan 68° = 2.475 (given in tables). Always show the multiplication step clearly to avoid calculation mistakes.

Question 4. Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50 m.
Answer:
Let one person A be at a distance x and the second person B be at a distance of y from the foot of the tower.
Given that angle of elevation of A is 30°
\( \tan 30° = \frac{50}{x} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{50}{x} \)
∴ x = 50√3 = 86.60 m
The angle of elevation of B is 38°
\( \tan 38° = \frac{50}{y} \)
\( \Rightarrow 0.7813 = \frac{50}{y} \)
∴ y ≈ 64 m
So, distance between A and B is x + y = 150.6 m
In simple words: Two people stand on opposite sides of a tower. One is 86.6 m away, the other is 64 m away. Total distance between them is 150.6 m.

[Diagram: This diagram shows a tower in the middle with two people on opposite sides, showing their angles of elevation as 30° and 38°.]

📝 Teacher's Note: Draw the tower in the middle of the board. Mark the two people on opposite sides. This helps students see why we add the two distances.

🎯 Exam Tip: Remember tan 30° = 1/√3 and use tan 38° = 0.7813 from tables. Show both calculations separately, then add the distances.

Question 5. A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30° with the ground.
Answer:
Let the length of the rope be x m.
Now,
\( \sin 30° = \frac{60}{x} \)
\( \Rightarrow \frac{1}{2} = \frac{60}{x} \)
∴ x = 120 m
So, the length of the rope is 120m.
In simple words: The kite is 60 m high. The string makes 30° with ground. Using sine ratio, the string length is 120 m.

[Diagram: This diagram shows a right triangle with the kite string as hypotenuse, height 60 m as vertical side, and 30° angle with ground.]

📝 Teacher's Note: This problem uses sine, not tangent. Show students when to use sin (when we know height and hypotenuse) vs tan (when we know height and base).

🎯 Exam Tip: Remember sin 30° = 1/2. Write this clearly. The string is always the longest side (hypotenuse) in these problems.

Question 6. A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45°, (ii) 60°. Find the height of the tower in each case.
Answer:
Let the height of the tower be h m.
(i) Here θ = 45°
∴ \( \tan 45° = \frac{h - 1.6}{20} \)
\( \Rightarrow 1 = \frac{h - 1.6}{20} \)
∴ h = 21.6 m
So, height of the tower is 21.6 m.
(ii) Here θ = 60°
∴ \( \tan 60° = \frac{h - 1.6}{20} \)
\( \Rightarrow \sqrt{3} = \frac{h - 1.6}{20} \)
∴ h = 20 × 1.732 + 1.6 = 36.24 m
So, height of the tower is 36.24 m.
In simple words: The boy's eyes are 1.6 m above ground. So we subtract this from total tower height. Case (i): tower is 21.6 m. Case (ii): tower is 36.24 m.

📝 Teacher's Note: This is tricky because the boy has height too. The angle is from his eye level, not ground level. Always subtract the observer's height.

🎯 Exam Tip: Write "h - 1.6" in the numerator, not just "h". This shows you understand the boy's height matters. Remember tan 45° = 1 and tan 60° = √3.

Question 7. The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?
Answer:
Let the height of the tree after breaking be h m.
Here θ = 45°
∴ \( \tan 45° = \frac{h}{15} \)
\( \Rightarrow 1 = \frac{h}{15} \)
∴ h = 15 m
Now, length of the tree broken by the wind = \( \frac{15}{\sin 45°} = 15\sqrt{2} = 21.21 \) m
So, height of the tree before it was broken is (15 + 21.21) m = 36.21 m.
In simple words: The tree broke and bent down. The bottom part is 15 m high. The broken part is 21.21 m long. Total original height was 36.21 m.

📝 Teacher's Note: Draw this carefully. The tree has two parts - the standing part (15 m) and the bent part (21.21 m). The bent part touches ground 15 m away from the root.

🎯 Exam Tip: Find the standing height first using tan. Then find the bent part length using sin. Add both parts for total original height.

Question 8. The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?
Answer:
Let AB be the unfinished tower and C be the top of the tower when finished. Let P be a point 80 m from the foot A.
In △ BAP,
\( \tan 30° = \frac{AB}{AP} \)
\( \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{80} \)
\( \Rightarrow AB = \frac{80}{\sqrt{3}} = 46.19 \) m
In △ CAP,
\( \tan 60° = \frac{AC}{AP} \)
\( \Rightarrow \sqrt{3} = \frac{AC}{80} \)
\( \Rightarrow AC = 80\sqrt{3} = 138.56 \) m
Therefore, the tower must be raised by (138.56 - 46.19) m = 92.37 m
In simple words: The unfinished tower is 46.19 m tall. When finished, it should be 138.56 m tall. So it needs to be raised by 92.37 m more.

[Diagram: This diagram shows a tower with two heights - current height AB and final height AC, with angles 30° and 60° from point P which is 80 m away.]

📝 Teacher's Note: This problem has two separate calculations. First find current height using 30°. Then find final height using 60°. Subtract to get the difference.

🎯 Exam Tip: Calculate both heights separately. Show the subtraction clearly. Write "must be raised by" in your final answer to match the question wording.

Question 9. At a particular time, when the sun's altitude is 30°, the length of the shadow of a vertical tower is 45 m. Calculate
(i) the length of the tower.
(ii) the length of the shadow of the same tower, when the sun's altitude is
(a) 45° (b) 60°
Answer:
Let the length of the tower be h m.

(i) Here \( \theta = 30° \)
\( \tan 30° = \frac{h}{45} \)
\( \implies \frac{1}{\sqrt{3}} = \frac{h}{45} \)
\( \implies h = 25.98 \text{ m} \)
Hence the length of the tower is 25.98 m.

(ii) Let the length of the shadow be x m.
(a) Here, \( \theta = 45° \)
\( \tan 45° = \frac{25.98}{x} \)
\( \implies 1 = \frac{25.98}{x} \)
\( \implies x = 25.98 \text{ m} \)
Hence the length of the shadow is 25.98 m

(b) Here, \( \theta = 60° \)
\( \tan 60° = \frac{25.98}{x} \)
\( \implies \sqrt{3} = \frac{25.98}{x} \)
\( \implies x = \frac{25.98}{\sqrt{3}} = 15 \text{ m} \)
Hence the length of the shadow is 15 m.
In simple words: We use the sun's angle to find how tall the tower is. Then we use different sun angles to find how long the shadow will be at those times.

📝 Teacher's Note: Draw a right triangle on the board. Show how the tower height stays the same but shadow length changes when sun angle changes. Use a torch and a pencil to show this in class.

🎯 Exam Tip: Always write "tan angle = height/shadow" first. Then substitute the known values. Show all steps clearly for full marks.

Question 10. Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32°24′ with the pole and when it is turned to rest against another pole, it makes angle 32°24′ with the road. Calculate the width of the road.
Answer:

[Diagram: Shows two vertical poles P and Q on either side of a road. A ladder AB of length 30m is shown in two positions - first touching pole P making angle 32°24′ with the pole, then rotated to touch pole Q making angle 32°24′ with the ground.]

Let AB be the ladder and \( \angle ABP = 32°24′ \).
\( \frac{BP}{AB} = \sin 32°24′ \)
\( \implies BP = 30 \times 0.536 = 16.08 \)

When rotated, let the ladder be AC and \( \angle CAQ = 32°24′ \).
\( \frac{BQ}{BC} = \cos 32°24′ \)
\( \implies BQ = 30 \times 0.844 = 25.32 \)
Hence, width of the road = (16.08 + 25.32) = 41.4 m
In simple words: The ladder touches two poles at different angles. We find how far each touch point is from the road center, then add them to get total road width.

📝 Teacher's Note: Use two chairs as poles and a stick as ladder. Show students how the ladder position changes but length stays same. This makes the problem very clear.

🎯 Exam Tip: Draw the diagram carefully. Mark all angles and distances clearly. Remember to add both distances to get total width.

Question 11. Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?
Answer:

[Diagram: Shows a vertical cliff with two climbers A and B at different heights. Observer C is 40m away from the cliff base P on level ground, looking up at angles 48° and 57°.]

Let P be the foot of the cliff on level ground.
Then, \( \angle ACP = 48° \) and \( \angle BCP = 57° \)
\( \frac{BP}{PC} = \tan 57° \)
\( \implies BP = 40 \times 1.539 = 61.57 \text{ m} \)

Also, \( \frac{AP}{PC} = \tan 48° \)
\( \implies AP = 40 \times 1.110 = 44.4 \text{ m} \)
Hence, distance between the climbers = AB = BP - AP = 17.17 m
In simple words: We find the height of each climber from ground level. Then we subtract to find the distance between them on the cliff face.

📝 Teacher's Note: Use a wall and two students at different heights. Show how observer angle changes with height. Draw this clearly on board with all measurements marked.

🎯 Exam Tip: Find both heights first, then subtract. Always write "distance between climbers = BP - AP" for clear working. Show all calculations step by step.

Question 12. A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28° and the angle of depression of the bottom of the pole is 13°. Calculate the height of the pole.
Answer:

[Diagram: Shows a flag pole PQ with point P at top and Q at bottom. A man at point A stands 9m away, looking up at 28° to top P and down at 13° to bottom Q.]

Let AB be the man and PQ be the flag-pole.
Given, AR = 9 m.
Also, \( \angle PAR = 28° \) and \( \angle QAR = 13° \)
\( \frac{PR}{AR} = \tan 28° \)
\( \implies PR = 9 \times 0.532 = 4.788 \text{ m} \)

Also, \( \frac{RQ}{AR} = \tan 13° \)
\( \implies RQ = 9 \times 0.231 = 2.079 \text{ m} \)
Hence, height of the pole = PR + RQ = 6.867 m
In simple words: The pole has two parts - one above eye level and one below. We find both parts separately using the angles, then add them together.

📝 Teacher's Note: Show students that flag poles often go into ground or have base below eye level. Use elevation and depression angles to find total height from ground to top.

🎯 Exam Tip: Draw horizontal line at eye level. Mark elevation angle above and depression angle below. Find both distances and add them for total height.

Question 13. From the top of a cliff 92 m high, the angle of depression of a buoy is 20°. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.
Answer:

[Diagram: Shows a cliff AB of height 92m. From top A, the angle of depression to buoy C is 20°. B is at the foot of the cliff.]

Let AB be the cliff and C be the buoy.
Given, AB = 92 m.
Also, \( \angle ACB = 20° \)
\( \frac{AB}{BC} = \tan 20° \)
\( \implies BC = \frac{92}{0.3640} = 252.7 \text{ m} \approx 253 \text{ m} \)
Hence, the buoy is at a distance of 253 m from the foot of the cliff.
In simple words: We look down from cliff top to the buoy. Using the angle and cliff height, we find how far the buoy is from the cliff bottom.

📝 Teacher's Note: Explain that depression angle is measured downward from horizontal. Draw horizontal line from cliff top, then show angle going down to the buoy.

🎯 Exam Tip: Use "tan depression angle = height/distance" formula. Always round to nearest metre when asked. Show the rounding step clearly.

Exercise 22 B

Question 1. In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.
Answer:

[Diagram: Shows a right triangle with AB perpendicular to BD. Point C is on BD with DC = 30m. Angles marked are ∠ADB = 30° and ∠ACB = 45°.]

In △ABC,
\( \frac{AB}{BC} = \tan 45° = 1 \)
\( \implies BC = AB = X \)

In △ABD,
\( \frac{AB}{BD} = \tan 30° \)
\( \frac{X}{30 + X} = \frac{1}{\sqrt{3}} \)
\( \implies 30 = \left(\sqrt{3} - 1\right) X \)
\( \therefore X = \frac{30}{1.732 - 1} = 40.98 \text{ m} \)
In simple words: We use two triangles with known angles. First triangle gives us a simple relation, second triangle helps us find the exact value.

📝 Teacher's Note: Show that 45° gives tan = 1, so opposite = adjacent. Use this simple relation first, then apply it to the more complex triangle.

🎯 Exam Tip: Always use the simpler angle (45°) first to get a relation. Then substitute this in the equation from the other angle. Show all algebra steps clearly.

Question 2. Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60° to 30°.
Answer:

[Diagram: Shows a tree AB. Observer moves from point C to point D, walking 20m away. Angle of elevation changes from 60° at C to 30° at D.]

Let AB be the tree of height h m.
Let the two points be C and D such that CD = 20 m, ∠ADB = 30° and ∠ACB = 60°

In △ABC,
\( \frac{AB}{BC} = \tan 60° = \sqrt{3} \)
\( \implies BC = \frac{AB}{\sqrt{3}} = \frac{h}{\sqrt{3}} \)

In △ABD,
\( \frac{AB}{BD} = \tan 30° \)
\( \frac{h}{20 + BC} = \frac{1}{\sqrt{3}} \)
\( \implies \sqrt{3} h = 20 + \frac{h}{\sqrt{3}} \)
\( \therefore h = \frac{20}{\sqrt{3} - \frac{1}{\sqrt{3}}} = \frac{20}{1.154} = 17.32 \text{ m} \)

Hence, height of the tree is 17.32 m.
In simple words: When we move away from a tree, the angle to look up gets smaller. We use both positions and angles to find the tree height.

📝 Teacher's Note: Take students outside and actually measure angles to a tall object from two distances. This makes the concept very real and easy to understand.

🎯 Exam Tip: Set up two equations - one from each position. Solve by substitution. Always check that the farther position has the smaller angle.

Question 3. Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.
Answer:
In simple words: When we walk closer to a building, the angle to look up gets bigger. We use both positions and angles to find the building height.

📝 Teacher's Note: This is similar to Question 2 but we walk towards instead of away. Show how angles increase when we get closer to tall objects.

🎯 Exam Tip: Draw diagram carefully showing movement direction. When walking towards, the closer position has the larger angle. Set up two equations and solve.

Question 4. From the top of a light house 100 m high, the angles of depression of two ships are observed as 48° and 36° respectively. Find the distance between the two ships(in the nearest metre) if:
(i) the ships are on the same side of the light house.
(ii) the ships are on the opposite sides of the light house.
Answer:
Given:
Height of lighthouse AB = 100 m
Angle of depression to first ship = 48°
Angle of depression to second ship = 36°

Step 1: Find distance BC using angle of depression 48°.
In triangle ABC, \( \frac{AB}{BC} = \tan 48° \)
\( BC = \frac{100}{\tan 48°} = \frac{100}{1.1106} = 90.04 \) m

Step 2: Find distance BD using angle of depression 36°.
In triangle ABD, \( \frac{AB}{BD} = \tan 36° \)
\( BD = \frac{100}{\tan 36°} = \frac{100}{0.7265} = 137.64 \) m

Step 3: Calculate distance between ships.
(i) If the ships are on the same side of the light house:
Distance between ships = BD - BC = 137.64 - 90.04 = 48 m

(ii) If the ships are on the opposite sides of the light house:
Distance between ships = BD + BC = 137.64 + 90.04 = 228 m

Final Answer:
(i) Distance = 48 m
(ii) Distance = 228 m

In simple words: We use tan ratios to find how far each ship is from the bottom of the lighthouse. Then we add or subtract these distances based on which side the ships are on.

[Diagram: This diagram shows a lighthouse AB with height 100m, and two ships C and D at different positions. The angles of depression from the top of lighthouse to the ships are 48° and 36° respectively.]

📝 Teacher's Note: Draw the diagram clearly on the board. Show students that angle of depression from top equals angle of elevation from bottom. This makes the trigonometry easier to understand.

🎯 Exam Tip: Always draw a clear diagram first. Mark all angles and distances. Remember that same side means subtract distances, opposite sides means add them.

Question 5. Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30°; find the height of the pillars and the position of the point.
Answer:
Given:
Width of roadway BD = 150 m
Angle of elevation to pillar A = 60°
Angle of elevation to pillar C = 30°
Both pillars have equal height h

Step 1: Set up equations using trigonometry.
Let P be the point of observation
Let BP = x, then PD = (150 - x)

In triangle ABP: \( \frac{AB}{BP} = \tan 60° \)
\( \frac{h}{x} = \sqrt{3} \)
\( x = \frac{h}{\sqrt{3}} \)

Step 2: Use second triangle.
In triangle CDP: \( \frac{CD}{DP} = \tan 30° \)
\( \frac{h}{150-x} = \frac{1}{\sqrt{3}} \)
\( 150 - x = h\sqrt{3} \)

Step 3: Solve for height.
Substituting \( x = \frac{h}{\sqrt{3}} \):
\( 150 - \frac{h}{\sqrt{3}} = h\sqrt{3} \)
\( 150 = \frac{h}{\sqrt{3}} + h\sqrt{3} \)
\( 150 = h(\frac{1}{\sqrt{3}} + \sqrt{3}) \)
\( 150 = h(\frac{1 + 3}{\sqrt{3}}) = h(\frac{4}{\sqrt{3}}) \)
\( h = \frac{150\sqrt{3}}{4} = \frac{150 × 1.732}{4} = 64.95 \) m

Step 4: Find position of point.
\( x = \frac{h}{\sqrt{3}} = \frac{64.95}{1.732} = 37.5 \) m

Final Answer:
Height of pillars = 64.95 m
Position of point = 37.5 m from first pillar

In simple words: We use the fact that both pillars are the same height. We make equations using tan ratios and solve them together to find the height and position.

[Diagram: This diagram shows two equal pillars AB and CD on either side of a 150m wide road, with point P between them where angles of elevation are 60° and 30°.]

📝 Teacher's Note: Emphasize that the total distance BP + PD = 150m. This constraint is key to solving the problem. Show students how to set up two equations with two unknowns.

🎯 Exam Tip: Set up your equations carefully. Use the fact that both pillars have equal height. Always check that your distances add up to the total width.

Question 6. From the figure, given below, calculate the length of CD.
Answer:
Given:
CB = 15 m
Angle ADE = 22°
Angle ACB = 47°

Step 1: Find AE using triangle AED.
In triangle AED: \( \frac{AE}{DE} = \tan 22° \)
\( AE = DE × \tan 22° = 15 × 0.404 = 6.06 \) m

Step 2: Find AB using triangle ABC.
In triangle ABC: \( \frac{AB}{BC} = \tan 47° \)
\( AB = BC × \tan 47° = 15 × 1.072 = 16.09 \) m

Step 3: Calculate CD.
\( CD = BE = AB - AE = 16.09 - 6.06 = 10.03 \) m

Final Answer: CD = 10.03 m

In simple words: We find the heights of two triangles and subtract them to get the length CD. It's like finding the difference between two heights.

[Diagram: This diagram shows a rectangular shape with diagonal lines creating triangles, with angles 22° and 47° marked, and a base of 15m.]

📝 Teacher's Note: Help students see that CD equals the difference between two heights. Draw vertical lines clearly to show AB and AE. This makes the subtraction obvious.

🎯 Exam Tip: Find each height separately first. Then subtract to get the final answer. Show all steps clearly for full marks.

Question 7. The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find:
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Answer:
Given:
First angle of elevation = 60°
Second angle of elevation = 45° (from 30 m higher point)
Vertical distance between observation points = 30 m

Step 1: Set up the problem.
Let AB be the tower of height h
Let C and D be the two observation points with CD = 30 m
Let BC = horizontal distance = x

Step 2: Use triangle ADE.
In triangle ADE: \( \frac{AE}{DE} = \tan 45° = 1 \)
\( AE = DE = BC = x \)
So \( AE = x \)

Step 3: Use triangle ABC.
In triangle ABC: \( \frac{AB}{BC} = \tan 60° = \sqrt{3} \)
\( \frac{h}{x} = \sqrt{3} \)
\( h = x\sqrt{3} \)

Step 4: Use the relationship between heights.
\( AE + 30 = AB \)
\( x + 30 = h = x\sqrt{3} \)
\( x + 30 = x\sqrt{3} \)
\( 30 = x\sqrt{3} - x = x(\sqrt{3} - 1) \)
\( x = \frac{30}{\sqrt{3} - 1} = \frac{30}{1.732 - 1} = \frac{30}{0.732} = 40.98 \) m

Step 5: Find tower height.
\( h = x\sqrt{3} = 40.98 × 1.732 = 70.98 \) m

Final Answer:
(i) Height of tower = 70.98 m
(ii) Horizontal distance = 40.98 m

In simple words: We use two different viewing angles from different heights to create equations. Then we solve these equations to find both the tower height and horizontal distance.

[Diagram: This diagram shows a tower AB with two observation points C and D, where D is 30m above C, creating different angles of elevation 60° and 45°.]

📝 Teacher's Note: Draw the diagram step by step. Show students how the 30m vertical difference creates two different triangles. This helps them see why we get two equations.

🎯 Exam Tip: Set up your variables clearly (h for height, x for distance). Use the relationship that one observation point is 30m above the other. Show all algebraic steps.

Question 8. From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.
Answer:
Given:
Height of cliff = 60 m
Angle of depression to top of tower = 30°
Angle of depression to bottom of tower = 60°

Step 1: Set up the problem.
Let the cliff be represented by vertical line AC = 60 m
Let the tower be BD with height h
Let horizontal distance = x

Step 2: Use angle of depression to bottom of tower.
Angle ACB = 60° (angle of depression equals angle of elevation)
In triangle ABC: \( \frac{AC}{BC} = \tan 60° = \sqrt{3} \)
\( \frac{60}{x} = \sqrt{3} \)
\( x = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \) m

Step 3: Use angle of depression to top of tower.
Let E be on the same level as D (top of tower)
In triangle ADE: \( \frac{AE}{DE} = \tan 30° = \frac{1}{\sqrt{3}} \)
AE = AC - EC = 60 - h
DE = BC = x = 20\sqrt{3}

\( \frac{60 - h}{20\sqrt{3}} = \frac{1}{\sqrt{3}} \)
\( 60 - h = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \)
\( 60 - h = 20 \)
\( h = 40 \) m

Final Answer: Height of tower = 40 m

In simple words: We use the two angles of depression to find where the tower starts and ends. The difference between these two positions gives us the tower height.

[Diagram: This diagram shows a cliff of 60m height and a tower, with angles of depression 30° and 60° from the top of cliff to top and bottom of tower respectively.]

📝 Teacher's Note: Help students understand that angle of depression from top equals angle of elevation from bottom. Draw horizontal lines to make the angles clear.

🎯 Exam Tip: Always convert angles of depression to angles of elevation for easier calculation. Set up two separate triangles and solve step by step.

Question 9. A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60°. Assuming that the boat sails at a uniform speed, determine:
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Answer:

[Diagram: This diagram shows a cliff AB with a man on top observing a boat at two different positions. The cliff height is shown, and the angles of depression are marked as 30° and 60°.]

Given:
Height of cliff = 500 m
Initial angle of depression = 30°
Final angle of depression = 60°
Time between observations = 3 minutes

Step 1: Set up the problem using the diagram.
Let AB be the cliff and C, D be the two positions of the boat.
AB = 500 m, ∠ADE = 30° and ∠ACB = 60°

Step 2: Find BC using triangle ABC.
In △ABC,
\( \frac{AB}{BC} = \tan 60° = \sqrt{3} \)
\( BC = \frac{AB}{\sqrt{3}} = \frac{500}{\sqrt{3}} \) m

Step 3: Find DE using triangle ADE.
In △ADE,
\( \frac{AE}{DE} = \tan 30° \)
\( AE = DE \tan 30° = \frac{DE}{\sqrt{3}} \)
Since AE = AB = 500 m,
\( DE = 500\sqrt{3} \) m

Step 4: Find distance CD traveled in 3 minutes.
\( CD = DE - BC = 500\sqrt{3} - \frac{500}{\sqrt{3}} \)
\( CD = 500\sqrt{3} - \frac{500\sqrt{3}}{3} = \frac{1500\sqrt{3} - 500\sqrt{3}}{3} = \frac{1000\sqrt{3}}{3} \) m

Step 5: Find speed of boat.
Speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{1000\sqrt{3}/3}{3} = \frac{1000\sqrt{3}}{9} \) m/min
Speed = \( \frac{1000 × 1.732}{9} = 192.45 \) m/min
Speed = \( \frac{192.45}{60} = 3.21 \) m/sec

Step 6: Find time to reach shore from position C.
Distance BC = \( \frac{500}{\sqrt{3}} = \frac{500}{1.732} = 288.67 \) m
Time = \( \frac{288.67}{192.45} = 1.5 \) minutes

(i) Time to reach shore = 1.5 minutes
(ii) Speed of boat = 3.21 m/sec
In simple words: The boat moves closer to the cliff. We use the angle changes to find how far it moved in 3 minutes. Then we can find its speed and remaining time.

📝 Teacher's Note: Draw the diagram clearly on the board. Show students that angle of depression is measured downward from horizontal. Use tan ratios to find distances.

🎯 Exam Tip: Always draw and label the diagram first. Write "Given" and list all values. Show each step clearly with proper units in the final answer.

Question 10. A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.
Answer:

[Diagram: This diagram shows a lighthouse AB and two positions of a boat C and D. The lighthouse height is marked as 150 m, and angles of elevation are shown as 60° and 45°.]

Given:
Height of lighthouse AB = 150 m
Initial angle of elevation = 60°
Final angle of elevation = 45°
Time taken = 2 minutes

Step 1: Find BC using triangle ABC.
In △ABC,
\( \frac{AB}{BC} = \tan 60° = \sqrt{3} \)
\( BC = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3} \) m

Step 2: Find BD using triangle ABD.
In △ABD,
\( \frac{AB}{BD} = \tan 45° = 1 \)
\( BD = 150 \) m

Step 3: Find distance CD traveled.
\( CD = BD - BC = 150 - 50\sqrt{3} \)
\( CD = 150 - 50 × 1.732 = 150 - 86.6 = 63.4 \) m

Step 4: Find speed of boat.
Speed = \( \frac{\text{Distance}}{\text{Time}} = \frac{63.4}{2} = 31.7 \) m/min
Speed = \( \frac{31.7}{60} = 0.53 \) m/sec

Speed of boat = 0.53 m/sec
In simple words: The boat moves away from the lighthouse. The angle gets smaller as it moves away. We find the distance moved in 2 minutes to get the speed.

📝 Teacher's Note: Explain that angle of elevation decreases as the boat moves away. Use real examples like looking up at a tall building - the angle gets smaller as you walk away.

🎯 Exam Tip: Remember tan 45° = 1 and tan 60° = √3. Convert final answer to m/sec by dividing by 60. Show all conversion steps.

Question 11. A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find:
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Answer:

[Diagram: This diagram shows a tree AB on one side of a river, with a person at two positions C and D. The distance CD is 40 m, and angles of elevation are marked as 60° and 30°.]

Given:
Initial angle of elevation = 60°
Final angle of elevation = 30°
Distance moved away from bank = 40 m

Let height of tree = h and width of river = w

Step 1: Use triangle ABC (first position).
\( \frac{AB}{BC} = \tan 60° = \sqrt{3} \)
\( \frac{h}{w} = \sqrt{3} \)
\( h = w\sqrt{3} \) ... (1)

Step 2: Use triangle ABD (second position).
\( \frac{AB}{BD} = \tan 30° = \frac{1}{\sqrt{3}} \)
\( \frac{h}{w + 40} = \frac{1}{\sqrt{3}} \)
\( h = \frac{w + 40}{\sqrt{3}} \) ... (2)

Step 3: Solve equations (1) and (2).
From (1) and (2):
\( w\sqrt{3} = \frac{w + 40}{\sqrt{3}} \)
\( 3w = w + 40 \)
\( 2w = 40 \)
\( w = 20 \) m

Step 4: Find height of tree.
\( h = w\sqrt{3} = 20\sqrt{3} = 20 × 1.732 = 34.64 \) m

(i) Height of tree = 34.64 m
(ii) Width of river = 20 m
In simple words: As the person moves away from the river, the angle to see the tree top gets smaller. We use both angles to find the tree height and river width.

📝 Teacher's Note: Draw the river clearly with the tree on one side and person on the other. Show how moving away changes the angle. This is a typical two-angle problem.

🎯 Exam Tip: Set up two equations using the two triangles. Solve them step by step. Round the final answer to 2 decimal places as asked.

Question 12. The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45°. Find the height of the first tower.
Answer:

[Diagram: This diagram shows two towers AB and CD with horizontal distance 75 m between them. The second tower is 160 m high, and the angle of depression from its top to the first tower's top is 45°.]

Given:
Height of second tower AB = 160 m
Horizontal distance BD = 75 m
Angle of depression ∠ACE = 45°

Let height of first tower CD = h

Step 1: Use the angle of depression triangle ACE.
In △ACE,
\( \frac{AE}{EC} = \tan 45° = 1 \)
\( AE = EC = BD = 75 \) m

Step 2: Find height of first tower.
\( AE = AB - CD \)
\( 75 = 160 - h \)
\( h = 160 - 75 = 85 \) m

Height of first tower = 85 m
In simple words: From the top of the tall tower, we look down at the shorter tower. The 45° angle helps us find how much shorter it is.

📝 Teacher's Note: Show students that angle of depression is measured downward from horizontal. The horizontal distance equals EC in the right triangle formed.

🎯 Exam Tip: Remember tan 45° = 1. This makes the calculation very simple. Draw the horizontal line clearly to show the depression angle.

Question 13. The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun's altitude is 30° than when it was 45°. Prove that the height of the tower is y(√3 + 1) metres.
Answer:

[Diagram: This diagram shows a tower AB with two different shadow lengths. When sun's altitude is 45°, shadow is BD. When sun's altitude is 30°, shadow is BC. The difference CD = 2y metres.]

Given:
Sun's altitude changes from 45° to 30°
Shadow increases by 2y metres
To prove: Height = y(√3 + 1) metres

Let height of tower AB = h
Let shorter shadow BD = x (when angle = 45°)
Let longer shadow BC = x + 2y (when angle = 30°)

Step 1: Use triangle ABD (sun's altitude 45°).
\( \frac{AB}{BD} = \tan 45° = 1 \)
\( h = x \)
So BD = h

Step 2: Use triangle ABC (sun's altitude 30°).
\( \frac{AB}{BC} = \tan 30° = \frac{1}{\sqrt{3}} \)
\( \frac{h}{h + 2y} = \frac{1}{\sqrt{3}} \)
\( h\sqrt{3} = h + 2y \)
\( h(\sqrt{3} - 1) = 2y \)

Step 3: Solve for h.
\( h = \frac{2y}{\sqrt{3} - 1} \)

Multiply numerator and denominator by (√3 + 1):
\( h = \frac{2y(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2y(\sqrt{3} + 1)}{3 - 1} = \frac{2y(\sqrt{3} + 1)}{2} \)

\( h = y(\sqrt{3} + 1) \)

Hence proved: Height of tower = y(√3 + 1) metres
In simple words: As the sun gets lower, shadows get longer. We use the two different shadow lengths to find the tower height in terms of y.

📝 Teacher's Note: Explain that sun's altitude is the angle of elevation of the sun. Use a stick in sunlight to show how shadow length changes with sun's position.

🎯 Exam Tip: Remember to rationalize the denominator by multiplying by (√3 + 1). This is a "prove that" question, so you must show all algebraic steps clearly.

Question 14. An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.
Answer:
Let A be the aeroplane and B be the observer on the ground. The vertical height will be AC = 1 km = 1000 m. After 10 seconds, let the aeroplane be at point D.
Let the speed of the aeroplane be x m/sec.
∴ CE = 10x

[Diagram: Shows a triangle with observer B at ground level, aeroplane positions A and D at height 1000m, with angles of elevation 60° and 30° respectively, and horizontal distance CE = 10x m]

Step 1: In ΔABC,
\( \frac{AC}{BC} = \tan 60° \)
\( \Rightarrow \frac{1000}{BC} = \sqrt{3} \)
\( \Rightarrow BC = \frac{1000}{\sqrt{3}} \) m

Step 2: In ΔBDE,
\( \frac{DE}{BE} = \tan 30° \)
\( \Rightarrow BE = 1000\sqrt{3} \)
∴ CE = BE - BC

Step 3: Calculate distance traveled by aeroplane.
\( 10x = 1000\sqrt{3} - \frac{1000}{\sqrt{3}} \)
\( \Rightarrow x = 100\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = 100 \times 1.154 \)
= 115.4 m/sec
= \( 115.4 \times \frac{18}{5} \) km/hr = 415.67 km/hr

Hence, speed of the aeroplane is 415.67 km/hr.
In simple words: We use two triangles to find where the plane was first and second. Then we find how far it moved in 10 seconds. Speed = distance ÷ time.

📝 Teacher's Note: Draw the diagram first. Show students that angle of elevation gets smaller as the plane moves away. Use tan ratio to find distances.

🎯 Exam Tip: Always convert final answer to km/hr by multiplying m/s by 18/5. Show all steps clearly with proper triangle names.

Question 15. From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of the hill.
Answer:
Let AB be the hill of height 'h' km and C and D be the two consecutive stones such that CD = 1 km, ∠ACB = 30° and ∠ADB = 45°.

[Diagram: Shows a hill AB with height h, and two stones C and D at ground level, with angles of depression 30° and 45° from point A]

Step 1: In ΔABD,
\( \frac{AB}{BD} = \tan 45° = 1 \)
\( \Rightarrow BD = h \)

Step 2: In ΔABC,
\( \frac{AB}{BC} = \tan 30° \)
\( \Rightarrow \frac{h}{BC} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow BC = h\sqrt{3} \)

Step 3: Since CD = 1 km,
\( \frac{h}{\sqrt{3} - 1} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow h = \frac{\sqrt{3} + 1}{2} = \frac{2.732}{2} = 1.366 \) km

∴ BD = 1.366 km
∴ BC = BD + DC = 1.366 + 1 = 2.366 km

Hence, the two stones are at a distance of 1.366 km and 2.366 km from the foot of the hill.
In simple words: We use angles of depression to find how far each stone is from the bottom of the hill. The stones are 1 km apart from each other.

📝 Teacher's Note: Explain that angle of depression from top equals angle of elevation from bottom. Draw both triangles clearly on the board.

🎯 Exam Tip: Write the relationship CD = BC - BD = 1 km. This helps you set up the equation correctly. Always state final distances clearly.

Exercise 22 C

Question 1. Find AD:
(i)
[Diagram: Shows a rectangle BCDE with height 5m and width 20m, with point A above forming angles, where angle B is 32°]

(ii)
[Diagram: Shows triangle with base BC, point A above with 30m marked, and angle of 48° at point C]

Answer:
(i) In ΔAEB,
\( \frac{AE}{BE} = \tan 32° \)
\( \Rightarrow AE = 20 \times 0.6249 = 12.50 \) m
∴ AD = AE + ED = 12.50 + 5 = 17.50m

(ii) In ΔABC,
∠ACD = ∠ABC + ∠BAC
and ∠ABC = ∠BAC (∵ AC = BC)
∴ ∠ABC = ∠BAC = \( \frac{48°}{2} = 24° \)

Now, \( \frac{AD}{AB} = \sin 24° \)
\( \Rightarrow AD = 30 \times 0.4067 = 12.20 \) m
In simple words: We use trigonometry to find the height. In part (i), we add the rectangle height to the triangle height.

📝 Teacher's Note: For part (ii), show students that when two sides are equal, the base angles are also equal. Use the exterior angle property.

🎯 Exam Tip: Draw perpendiculars clearly. Label all known measurements first. In isosceles triangles, base angles are equal.

Question 2. In the following diagram, AB is a floor-board; PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.
[Diagram: Shows a cubical box PQRS on floor with board AB at 60° angle]

Answer:
Step 1: In ΔPSB,
\( \frac{PS}{PB} = \sin 60° \)
\( \Rightarrow PB = \frac{2}{\sqrt{3}} = 1.155 \) m

Step 2: In ΔAPQ,
∠APQ = 60°
\( \frac{PQ}{AP} = \cos 60° \)
\( \Rightarrow AP = \frac{1}{1/2} = 2 \) m

Step 3: Calculate total length.
∴ AB = AP + PB = 2 + 1.155 = 3.155 m
In simple words: The board touches the cube and makes a 60° angle. We find the two parts of the board separately and add them.

📝 Teacher's Note: Use a real box and a stick to show this setup. The cube edge is always 1m - this is given information.

🎯 Exam Tip: Break the problem into two right triangles. Use sin for opposite/hypotenuse and cos for adjacent/hypotenuse. Add the segments carefully.

Question 3. Calculate BC.
[Diagram: Shows triangle with point A at top, base BC, point D with vertical line showing 20m height, angles marked as 35° at B and 42° at A]

Answer:
Step 1: In ΔADC,
\( \frac{CD}{AD} = \tan 42° \)
\( \Rightarrow CD = 20 \times 0.9004 = 18.008 \) m

Step 2: In ΔADB,
\( \frac{AD}{BD} = \tan 35° \)
\( \Rightarrow BD = \frac{AD}{\tan 35°} = \frac{20}{0.7002} = 28.563 \) m

Step 3: Calculate BC.
∴ BC = BD - CD = 28.563 - 18.008 = 10.55 m
In simple words: We find two distances from the vertical line to points B and C. The difference between them gives us BC.

📝 Teacher's Note: Show students that we need to subtract the smaller distance from the larger one. Draw the perpendicular clearly.

🎯 Exam Tip: Always identify which triangle contains which angle. Use tan ratio when you have opposite and adjacent sides. Check if your final answer makes sense.

Question 4. Calculate AB.
Answer:
[Diagram: This diagram shows a triangle with points A, M, and B forming the base. From M, there are two lines going up - one to the left making a 30° angle with the base (length 6 m), and one to the right at 47° (length 5 m). These lines meet the top edge of a rectangular shape.]

Step 1: In triangle AMOB
cos 30° = \( \frac{AO}{MO} \)
\( \frac{\sqrt{3}}{2} = \frac{AO}{6} \)
\( \implies AO = 5.20 \text{ m} \)

Step 2: In triangle BNO
sin 47° = \( \frac{OB}{NO} \)
\( 0.73 = \frac{OB}{5} \)
\( \implies OB = 3.65 \text{ m} \)

Step 3: Calculate total length
AB = OA + OB
\( \implies AB = 5.20 + 3.65 \)
\( \implies AB = 8.85 \text{ m} \)

In simple words: We broke the big line AB into two smaller parts. We used trigonometry (sine and cosine) to find each part. Then we added them together.

📝 Teacher's Note: Draw a right triangle on the board. Show students how cosine gives us the side next to the angle. Sine gives us the side opposite to the angle. Practice with simple triangles first.

🎯 Exam Tip: Always draw a clear diagram first. Label all sides and angles. Write "Given" and "To find" clearly. Show each trigonometric step separately for full marks.

Question 5. The radius of a circle is given as 15 cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, calculate: (i) the length of AB; (ii) the distance of AB from the centre C.
Answer:
[Diagram: This diagram shows a circle with center C. Chord AB is drawn with radii CA and CB both 15 cm long. The angle ACB is 131°. A perpendicular line CP is drawn from center C to chord AB.]

Given: CA = CB = 15 cm, ∠ACB = 131°
Drop a perpendicular CP from centre C to the chord AB.
Then CP bisects ∠ACB as well as chord AB.

∠ACP = 65.5°

Step 1: In triangle ACP
\( \frac{AP}{AC} = \sin(65.5°) \)
\( \implies AP = 15 \times 0.91 = 13.65 \text{ cm} \)

(i) AB = 2 AP = 2 × 13.65 = 27.30 cm

(ii) CP = AP cos(65.5°)
= 15 × 0.415 = 6.22 cm

In simple words: When we drop a line from the center to a chord, it cuts the chord into two equal parts. We used sine to find half the chord length, then doubled it. We used cosine to find the distance from center to chord.

📝 Teacher's Note: Use a compass to draw a circle on the board. Show students that the perpendicular from center to chord always bisects the chord. This is a key circle property.

🎯 Exam Tip: Always remember that perpendicular from center bisects the chord. Write this property clearly. Use sin for opposite side and cos for adjacent side in the right triangle formed.

Question 6. At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle is found to be 3/4. Find the height of the tower.
Answer:
[Diagram: This diagram shows a vertical tower AB with two observation points C and D on the ground. The distance CD is 192 m. From point C, the angle of elevation is θ where tan θ = 5/12. From point D, the angle of elevation is α where tan α = 3/4.]

Given: tan θ = \( \frac{5}{12} \)

\( \implies \frac{AB}{BC} = \frac{5}{12} \)

\( \implies AB = \frac{5}{12}BC \) ... (i)

Also: tan α = \( \frac{3}{4} \)

\( \implies \frac{AB}{BD} = \frac{3}{4} \)

\( \implies \frac{\frac{5}{12}BC}{BD} = \frac{3}{4} \)

\( \implies \frac{192 + BD}{BD} = \frac{3}{4} \times \frac{12}{5} \)

\( \implies BD = 240 \text{ m} \)

BC = (192 + 240) = 432 m

By (i), AB = \( \frac{5}{12} \times 432 = 180 \text{ m} \)

Hence, the height of the tower is 180 m.

In simple words: We had two different viewing positions. From each position, we got a different angle to look up at the tower. We used both angles to set up equations and solve for the tower height.

📝 Teacher's Note: Draw a simple tower and two stick figures looking up at it from different distances. Show how the angle gets bigger when you move closer. This helps students understand angle of elevation.

🎯 Exam Tip: Set up two separate equations using tan θ = height/base from both positions. Remember that the person moved 192 m closer, so subtract this from the original distance.

Question 7. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and at the top of the flagstaff is β. Prove that the height of the tower is \( \frac{h \tan α}{\tan β - \tan α} \).
Answer:
[Diagram: This diagram shows a vertical tower AB with a flagstaff AD on top. From point C on the ground, the angle of elevation to B (bottom of flagstaff) is α, and to A (top of flagstaff) is β. The flagstaff height AD = h.]

Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that ∠ACB = α, ∠DCB = β and AD = h.

In triangle ABC:
\( \frac{AB}{BC} = \tan α \)

\( \implies BC = \frac{x}{\tan α} \) ... (i)

In triangle DBC:
\( \frac{BD}{BC} = \tan β \)

\( \implies BD = \frac{x}{\tan α} \tan β \) [From (i)]

\( \implies (h + x) \tan α = x \tan β \)

\( \implies x \tan β - x \tan α = h \tan α \)

\( \implies x = \frac{h \tan α}{\tan β - \tan α} \)

Hence, height of the tower is \( \frac{h \tan α}{\tan β - \tan α} \)

In simple words: We have a tower with a flag pole on top. From one point on ground, we can see both the bottom and top of the flag pole at different angles. We used both angles to find the tower height.

📝 Teacher's Note: Use building blocks to show a tower with a small flag on top. Point out that the angle to the top of flag is bigger than the angle to bottom of flag. This makes the concept clear.

🎯 Exam Tip: Set up two right triangles - one for the tower height and one for the total height (tower + flagstaff). Use substitution carefully to eliminate the base distance and prove the formula.

Question 8. With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = 2/5. Calculate: (i) the distance AB in metres; (ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.
Answer:
[Diagram: This diagram shows a vertical pole BC of height 10 m. A man stands at point A, with his eye level 2 m above ground at point D. The angle of elevation from D to C is x°. Points are arranged to show the geometric setup.]

Let AD be the height of the man, AD = 2 m.
CE = (10 - 2) = 8 m

(i) In triangle CED,
\( \frac{CE}{DE} = \tan x° = \frac{2}{5} \)

\( \implies \frac{8}{DE} = \frac{2}{5} \)

\( \implies DE = 20 \text{ m} \)

Here AB = DE.
∴ AB = 20 m

(ii) Let A'D' be the new position of the man and θ be the angle of elevation of the top of the tower.
So, D'E = 15 m
In triangle CED,
\( \tan θ = \frac{CE}{D'E} = \frac{8}{15} = 0.533 \)

\( \implies θ = 28° \)

In simple words: The man's eye is 2 m above ground, so we only need to consider the height difference between his eye and the pole top. We used tan ratio to find horizontal distance in both cases.

📝 Teacher's Note: Show students that we consider the eye level of the person, not ground level. Draw a horizontal line from the person's eye to make a right triangle. This avoids confusion about heights.

🎯 Exam Tip: Always subtract the person's height from the total pole height when calculating angle of elevation. Use tan ratio = opposite/adjacent. Round final angles to nearest degree as asked.

Question 9. The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same line are complementary. Prove that the height of the tower is \( \sqrt{ab} \) metre.
Answer:
[Diagram: This diagram shows a vertical tower with two observation points at distances a and b from the base. The angles of elevation from these points are complementary (add up to 90°).]

Let PQ be the tower of height h metres.
Let A and B be two points on the ground such that PA = a metres and PB = b metres.
Let ∠PAQ = θ and ∠PBQ = (90° - θ) [complementary angles]

In right triangle PAQ:
\( \tan θ = \frac{h}{a} \)

In right triangle PBQ:
\( \tan (90° - θ) = \frac{h}{b} \)

Since tan(90° - θ) = cot θ = \( \frac{1}{\tan θ} \)

\( \implies \frac{1}{\tan θ} = \frac{h}{b} \)

\( \implies \frac{1}{\frac{h}{a}} = \frac{h}{b} \)

\( \implies \frac{a}{h} = \frac{h}{b} \)

\( \implies h^2 = ab \)

\( \implies h = \sqrt{ab} \)

Hence, the height of the tower is \( \sqrt{ab} \) metres.

In simple words: Two angles are complementary means they add up to 90°. We used the fact that tan of one angle equals cot of the other angle. This gave us a simple equation to solve.

📝 Teacher's Note: Show students that complementary angles have a special relationship: tan θ × tan(90° - θ) = 1. Use your hands to show two angles that add up to 90°. This helps remember the concept.

🎯 Exam Tip: Remember that tan(90° - θ) = cot θ = 1/tan θ. Set up two equations using tan ratio from both positions. Use the complementary angle relationship to solve and prove the formula.

Solution:

[Diagram: This diagram shows a triangle with vertices A, B, C, and D. Point A is at the top, B is at the bottom right, C is at the bottom left, and D is between C and B on the ground. The diagram shows angles α and β, and indicates measurements in metres.]

Let AB be the tower of height h metres.
Let C and D be two points on the level ground such that BC = b metres, BD = a metres, ∠ACB = α and ∠ADB = β.
Given, α + β = 90°
In ΔABC,
    \( \frac{AB}{BC} = \tan α \)
\( \Rightarrow \frac{h}{b} = \tan α \) -- (i)
In ΔABD,
    \( \frac{AB}{BD} = \tan β \)
\( \Rightarrow \frac{h}{a} = \tan(90° - α) = \cot α \) -- (ii)
Multiplying (i) by (ii), we get,
    \( \left(\frac{h}{a}\right)\left(\frac{h}{b}\right) = 1 \)
\( \Rightarrow h^2 = ab \)
\( \Rightarrow h = \sqrt{ab} \) metre
Hence, height of the tower is \( \sqrt{ab} \) metre.

📝 Teacher's Note: This is a classic height and distance problem using complementary angles. When two angles add up to 90°, their tan and cot are equal. This gives us the key relationship h² = ab.

🎯 Exam Tip: Always write "Given: α + β = 90°" first. Then use the fact that tan(90° - α) = cot α. The final answer h = √(ab) is what examiners look for.

Question 10. From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x° = \( \frac{5}{2} \) and the angle of depression of the foot D of the tower is y°, where tan y° = \( \frac{1}{4} \). Calculate the height CD of the tower in metres.

[Diagram: This diagram shows a rectangular building with point A at a window, point B at the base of the building, a tower CD, and point E at the same level as A. The diagram shows angles x° and y°.]

Answer:
Here, AB = DE = 10 m
In ΔADE,
    \( \frac{DE}{AE} = \tan y = \frac{1}{4} \)
\( \Rightarrow AE = 4DE = 4 × 10 = 40 \) m
In ΔAEC,
    \( \frac{CE}{AE} = \tan x = \frac{5}{2} \)
\( \Rightarrow CE = 40 × \frac{5}{2} = 100 \) m
∴ CD = DE + EC = (10 + 100) = 110 m
Hence, height of the tower CD is 110 m.
In simple words: From the window, we look down to see the bottom of the tower and up to see the top. We use both angles to find the total height step by step.

📝 Teacher's Note: Draw a clear diagram showing the window level. Students often forget that the tower height = height below window level + height above window level. Make this very clear.

🎯 Exam Tip: Always write CD = DE + EC in the final step. Show both triangles separately. Don't forget to add the two parts to get total tower height.

Question 11. A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

Answer:
Let AB be the tower of height 20 m.
Let θ be the angle of elevation of the top of the tower from point C.
Given, cos θ = 0.53
\( \Rightarrow θ = 58° \)
In ΔABC,
    \( \frac{AB}{BC} = \tan 58° \)
\( \Rightarrow \frac{20}{BC} = 1.6 \)
∴ \( BC = \frac{20}{1.6} = 12.5 \) m
In simple words: We know cos θ = 0.53, so θ = 58°. Then we use tan 58° = 1.6 to find the distance from the tower base.

[Diagram: This diagram shows a right triangle with tower AB of height 20 m, base BC, and angle θ at point C.]

📝 Teacher's Note: Students must remember that if cos θ is given, they need to find θ first, then use tan θ to get the horizontal distance. Calculator values for tan 58° ≈ 1.6.

🎯 Exam Tip: Write "Given: cos θ = 0.53" first. Then find θ = 58°. Use tan θ for height/base ratio. Always check if your calculator is in degree mode.

Question 12. A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate:
(i) the width of the river;
(ii) the height of the tree.

Answer:
Let AB be the tree and AC be the width of the river. Let D be a point such that CD = 50 m. Given that ∠BCA = 60° and ∠BDA = 30°.
In ΔBAD,
    \( \frac{BA}{AD} = \tan 30° \)
\( \Rightarrow BA = \frac{AD}{\sqrt{3}} \) -- (i)
In ΔBAC,
    \( \frac{BA}{AC} = \tan 60° \)
\( \Rightarrow BA = AC\sqrt{3} \) -- (ii)
From (i) and (ii), we get
    \( \frac{AD}{\sqrt{3}} = AC\sqrt{3} \)
\( \Rightarrow (50 + AC) = 3AC \)
∴ AC = 25 m
Thus, width of the river is 25 m.
From (ii),
    BA = 25 × 1.732 = 43.3 m
Hence, height of the tree is 43.3 m.
In simple words: The man moves back 50 m and the angle gets smaller. We use both positions to make two equations and solve for river width and tree height.

[Diagram: This diagram shows a tree BA on one side of a river, with the man at position C on the river bank, then at position D which is 50 m further away. Angles of 60° and 30° are shown.]

📝 Teacher's Note: This is the classic "two position" problem. Draw the diagram carefully showing the man's two positions. Explain that AD = AC + CD = AC + 50.

🎯 Exam Tip: Set up two separate triangles. Write tan 30° = 1/√3 and tan 60° = √3. Solve by substitution. Answer both parts clearly - (i) width = 25 m, (ii) height = 43.3 m.

Question 13. A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is 60° and the angle of elevation of the top of the pole, as seen from the foot of the tower is 30°. Find:
(i) the height of the tower;
(ii) the horizontal distance between the pole and the tower.

Answer:
Let AB be the tower and CD be the pole.
Given, CD = 20 m, ∠ADB = 60° and ∠CBD = 30°.
In ΔBDC,
    \( \frac{CD}{BD} = \tan 30° \)
\( \Rightarrow BD = 20\sqrt{3} \) m
In ΔDBA,
    \( \frac{AB}{BD} = \tan 60° = \sqrt{3} \)
\( \Rightarrow AB = 20\sqrt{3} × \sqrt{3} = 60 \) m
Hence,
(i) height of the tower = 60 m
(ii) horizontal distance between the pole and tower = 20 × 1.732 = 34.64 m
In simple words: We use the 20 m pole height and the two angles to find both the tower height and the distance between them.

[Diagram: This diagram shows tower AB and pole CD with the base distance BD. Angles of 30° and 60° are marked, and the distance of 50 m is shown.]

📝 Teacher's Note: This problem has two unknowns - tower height and distance. Use the pole height (given) with both angles to solve step by step. The key is that BD is the same in both triangles.

🎯 Exam Tip: Start with the triangle containing the known height (pole). Find BD first using tan 30°. Then use BD with tan 60° to find tower height AB. Show both parts clearly.

Question 14. A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. Find:
(i) the height of the tower, if the height of the pole is 20 m;
(ii) the height of the pole, if the height of the tower is 75 m.

Answer:
Let AB be the tower and CD be the pole.
Given, CD = 20 m, ∠ADB = 60° and ∠CBD = 30°
In ΔBDC,
    \( \frac{CD}{BD} = \tan 30° \)
\( \Rightarrow BD = 20\sqrt{3} \) m
In ΔDBA,
    \( \frac{AB}{BD} = \tan 60° = \sqrt{3} \)
\( \Rightarrow AB = 20\sqrt{3} × \sqrt{3} = 60 \) m
Hence,
(i) height of the tower = 60 m
(ii) horizontal distance between the pole and tower = 20 × 1.732 = 34.64 m
In simple words: From the top of the pole, we look up to see the tower top and down to see the tower bottom. We use both angles to find the tower height.

[Diagram: This diagram shows tower AB and pole CD, with point B at the base of the tower, point D at the top of the pole, and angles of 30° and 60° marked for depression and elevation respectively.]

📝 Teacher's Note: Draw the diagram carefully showing that both angles are measured from the horizontal line through the top of the pole. Depression angle goes downward, elevation angle goes upward.

🎯 Exam Tip: Remember that angle of depression to the tower base gives you the horizontal distance. Angle of elevation to the tower top gives you the tower height. Use both triangles formed at the pole top.

Question 15. From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and the angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.

[Diagram: Shows a point A at 36 m above lake surface, with angle of elevation 30° to bird B, and angle of depression 60° to bird's image B' in the water. The bird is at height h above lake surface.]

Answer:
Given:
Observer's height above lake = 36 m
Angle of elevation to bird = 30°
Angle of depression to bird's image = 60°

Step 1: Set up the problem.
Let A be the observer's position, B be the bird, B' be the bird's image in water.
Let BE = h m (height of bird above lake surface)

Step 2: Find relationship in triangle ABE.
\( \frac{BE}{AE} = \tan 30° \)
\( \frac{h}{AE} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow AE = \sqrt{3}h \) ... (i)

Step 3: Find relationship in triangle AB'E.
B'E = B'D + DE = 36 + 36 + h = 72 + h
\( \frac{B'E}{AE} = \tan 60° \)
\( \frac{72 + h}{AE} = \sqrt{3} \)
From (i): \( 72 + h = \sqrt{3} \times \sqrt{3}h = 3h \)

Step 4: Solve for h.
\( 72 + h = 3h \)
\( 72 = 2h \)
\( h = 36 \text{ m} \)

Step 5: Find actual height above lake surface.
Actual height = 36 + 36 = 72 m
In simple words: The observer sees the bird above and its mirror image below the water. Using angles and trigonometry, we find the bird flies 72 meters above the lake.

📝 Teacher's Note: Draw the diagram clearly showing the bird above water and its image below. Explain that the image appears as deep below water as the bird is high above it. This makes the calculation easier to understand.

🎯 Exam Tip: Always draw a clear diagram first. Mark all given angles and distances. The key is to use both triangles - one to the bird and one to its image. Show all steps clearly.

Question 16. A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.

[Diagram: Shows building AB with height h, man at two positions M and N, with angles 30° and 60° respectively, and MN = 60 m.]

Answer:
Given:
Initial angle of elevation = 30°
Final angle of elevation = 60°
Distance walked = 60 m

Step 1: Set up in triangle AMB.
\( \tan 30° = \frac{AB}{MB} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{60 + x} \)
\( \Rightarrow 60 + x = \sqrt{3}h \)
\( \Rightarrow x = \sqrt{3}h - 60 \) ... (1)

Step 2: Set up in triangle ANB.
\( \tan 60° = \frac{AB}{NB} \)
\( \sqrt{3} = \frac{h}{x} \)
\( \Rightarrow x = \frac{h}{\sqrt{3}} \) ... (2)

Step 3: Solve equations (1) and (2).
From (1) and (2):
\( \sqrt{3}h - 60 = \frac{h}{\sqrt{3}} \)
\( \Rightarrow 3h - 60\sqrt{3} = h \)
\( \Rightarrow 2h = 60\sqrt{3} \)
\( \Rightarrow h = 30\sqrt{3} \)

Step 4: Calculate final answer.
\( h = 30\sqrt{3} = 30 \times 1.732 = 51.96 \text{ m} \)
Height of building = 52 m (to nearest metre)
In simple words: The man walks closer to the building, so the angle becomes bigger. Using both positions and angles, we can find the building height is 52 meters.

📝 Teacher's Note: Show students that as you move closer to a tall object, the angle you look up increases. Use a simple example - looking at a tree from far away versus close up. This helps them visualize the problem.

🎯 Exam Tip: Set up two separate triangles and two equations. Always check your final answer makes sense - the building should be taller than the distance walked. Round to nearest metre as asked.

Question 17. As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of a light house in a horizontal line with its base, are 30° and 40° respectively. Find the distance between the two ships. Give your answer corrected to the nearest metre.

[Diagram: Shows lighthouse AB of height 80 m, with two ships at points D and C, angles of depression 30° and 40° respectively.]

Answer:
Given:
Height of lighthouse = 80 m
Angle of depression to ship D = 30°
Angle of depression to ship C = 40°

Step 1: Find distance to ship C.
In triangle ACB:
\( \tan 40° = \frac{80}{CB} \)
\( CB = \frac{80}{0.84} = 95.24 \text{ m} \)

Step 2: Find distance to ship D.
In triangle ADB:
\( \tan 30° = \frac{80}{DB} \)
\( DB = \frac{80}{0.58} = 137.93 \text{ m} \)

Step 3: Find distance between ships.
DC = DB - CB
x = 137.93 - 95.24
x = 42.69 ≈ 43 m
The distance between the two ships is 43 m.
In simple words: From the lighthouse top, we can see both ships. The farther ship has a smaller angle. We find each distance separately, then subtract to get the gap between ships.

📝 Teacher's Note: Explain that angles of depression are measured downward from horizontal. The farther the ship, the smaller the angle of depression. Use tan function because we have height and need horizontal distances.

🎯 Exam Tip: Draw the diagram with both angles clearly marked. Calculate each distance separately, then find the difference. Always round to nearest metre as asked. Check that the farther ship has the smaller angle.

Question 18. In the given figure, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find: (i) the horizontal distance between AB and CD. (ii) the height of the lamp post.

[Diagram: Shows building AB of 60 m height, lamp post CD, with angles of depression 30° to top D and 60° to bottom C.]

Answer:
Given:
Height of building AB = 60 m
Angle of depression to top of lamp post = 30°
Angle of depression to bottom of lamp post = 60°
Let BC = DE = x and CD = BE = y
Therefore, AE = AB - BE = 60 - y

(i) Finding horizontal distance:
In right triangle AED:
\( \tan 30° = \frac{AE}{DE} \)
\( \frac{1}{\sqrt{3}} = \frac{60 - y}{x} \)
\( \Rightarrow x = 60\sqrt{3} - y\sqrt{3} \) ... (1)

In right triangle ABC:
\( \tan 60° = \frac{AB}{BC} \)
\( \sqrt{3} = \frac{60}{x} \)
\( \Rightarrow x = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m} \)

(ii) Finding height of lamp post:
From equation (1):
\( 20\sqrt{3} = 60\sqrt{3} - y\sqrt{3} \)
\( y\sqrt{3} = 60\sqrt{3} - 20\sqrt{3} = 40\sqrt{3} \)
\( y = 40 \text{ m} \)

Final Answer:
(i) Horizontal distance = \( 20\sqrt{3} = 34.64 \text{ m} \)
(ii) Height of lamp post = 40 m
In simple words: We use two triangles - one to the top of the lamp post and one to the bottom. This gives us two equations to solve for both the distance and height.

📝 Teacher's Note: Show students how to label all parts clearly. The key insight is that AE is shorter than AB because we go from building top to lamp post top. Make sure students see the two separate right triangles.

🎯 Exam Tip: Always find the horizontal distance first using the bottom of the lamp post. Then use this distance in the second triangle to find the lamp post height. Show all working steps clearly and don't forget units.

Question 19. An aeroplane, at an altitude of 250 m, observes the angles of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.

[Diagram: This diagram shows a triangle with point A at the top representing the aeroplane at 250m altitude. Points B and C are on opposite banks of the river below, with angles of depression 60° and 45° respectively from point A.]

Answer:
Given:
Altitude of aeroplane = 250 m
Angle of depression to boat B = 60°
Angle of depression to boat C = 45°

Let A be the position of the airplane and let BC be the river. Let D be the point in BC just below the airplane. B and C are two boats on the opposite banks of the river with angles of depression 60° and 45° from A.

Step 1: In triangle ΔADC,
\( \tan 45° = \frac{AD}{DC} \)
\( 1 = \frac{250}{y} \)
\( y = 250 \text{ m} = DC \)

Step 2: In triangle ΔADB,
\( \tan 60° = \frac{AD}{BD} \)
\( \sqrt{3} = \frac{250}{x} \)
\( x = \frac{250}{\sqrt{3}} = \frac{250\sqrt{3}}{3} = \frac{250 \times 1.732}{3} = 144.3 \text{ m} = BD \)

Step 3: Width of river
BC = BD + DC = 144.3 + 250 = 394.3 ≈ 394 m

Width of the river = 394 m
In simple words: The aeroplane is above the middle of the river. We use tan ratios to find how far each boat is from the point below the plane. Then we add both distances to get the river width.

📝 Teacher's Note: Draw a big triangle on the board. Show students that angles of depression become angles of elevation in the triangle. This makes tan ratios easy to use.

🎯 Exam Tip: Always draw the diagram first. Mark all given angles and distances clearly. Write "Given" and list everything. Show each triangle calculation separately.

Question 20. The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answers correct to 3 significant figures.

[Diagram: This diagram shows two towers with horizontal distance 120m between them. From the top of the second tower, there's a 30° angle of elevation to the top of the first tower and a 24° angle of depression to the bottom.]

Answer:
Given:
Horizontal distance between towers = 120 m
Angle of elevation to top of first tower = 30°
Angle of depression to bottom of first tower = 24°

Step 1: In triangle ΔAEC,
\( \tan 30° = \frac{AE}{EC} \)
\( \frac{1}{\sqrt{3}} = \frac{AE}{120} \)
\( AE = \frac{120}{\sqrt{3}} = \frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} = 40 \times 1.732 = 69.28 \text{ m} \)

Step 2: In triangle ΔBEC,
\( \tan 24° = \frac{EB}{EC} \)
\( 0.4452 = \frac{EB}{120} \)
\( EB = 53.424 \text{ m} \)

Step 3: Height calculations
Height of first tower:
AB = AE + EB = 69.28 + 53.424 = 122.704 = 123 m (correct to 3 significant figures)

Height of second tower:
CD = EB = 53.424 m = 53.4 m (correct to 3 significant figures)

Height of first tower = 123 m
Height of second tower = 53.4 m
In simple words: From the top of the second tower, we look up to see the top of the first tower and look down to see its bottom. Using tan ratios, we find these distances and add them for the full height.

📝 Teacher's Note: Use two pencils of different lengths to show this problem. Hold them vertically and show how one observer looks up and down from the top of one pencil.

🎯 Exam Tip: Write the answer to 3 significant figures as asked. For 122.704, count from the first non-zero digit: 1-2-2 gives 123. Don't forget units (m).

Question 21. The angles of depression of two ships A and B as observed from the top of a light house 60m high, are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.

[Diagram: This diagram shows a lighthouse of height 60m with point T at the top. Ships A and B are on opposite sides, with angles of depression 60° and 45° respectively from point T.]

Answer:
Given:
Height of lighthouse = 60 m
Angle of depression to ship A = 60°
Angle of depression to ship B = 45°
Ships are on opposite sides of lighthouse

In the above figure:
OT = tower = 60m
A and B are the respective positions of ships

Step 1: In triangle ΔTAO
\( \tan 60° = \frac{TO}{AO} \)
\( \sqrt{3} = \frac{60}{AO} \)
\( AO = 20\sqrt{3} \text{ m} \)

Step 2: In triangle ΔTBO
\( \tan 45° = \frac{TO}{OB} \)
\( 1 = \frac{60}{OB} \)
\( OB = 60 \text{ m} \)

Step 3: Distance between ships
\( AB = AO + OB = 20\sqrt{3} + 60 = 20(1.732) + 60 = 94.64 ≈ 95 \text{ m} \)

Distance between the two ships = 95 m
In simple words: The lighthouse is in the middle. We find how far each ship is from the base of the lighthouse using tan ratios. Since ships are on opposite sides, we add both distances.

📝 Teacher's Note: Use a ruler standing on the desk as the lighthouse. Place two erasers on opposite sides to represent ships. This makes the concept very clear.

🎯 Exam Tip: When ships are on "opposite sides", always ADD the distances. If they were on the same side, you would subtract. Read the question carefully.