Samacheer Kalvi Class 10 Maths Solutions Chapter 6 Trigonometry Exercise 6.3

Question 1. From the top of a rock \( 50\sqrt{3} \) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer: Let the distance of the car from the rock be \( x \) meters.

In the right-angled triangle ABC: We know that the angle of depression is equal to the angle of elevation. So, \( \angle ACB = 30^\circ \).
We use the tangent function, which relates the opposite side to the adjacent side.
\( \tan 30^\circ = \frac{AB}{BC} \)
\( \frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{x} \)
Now, cross-multiply to solve for \( x \).
\( x = 50\sqrt{3} \times \sqrt{3} \)
\( x = 50 \times 3 \)
\( x = 150 \) m
So, the distance of the car from the base of the rock is 150 meters. Understanding the angle of depression is key here as it helps form the right triangle needed for calculation.
In simple words: We used the height of the rock and the angle given to find the distance of the car. The car is 150 meters away from the rock.

🎯 Exam Tip: Always draw a clear diagram for problems involving angles of elevation or depression to visualize the right-angled triangles and correctly identify the sides and angles.

Question 2. The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer: Let AD be the height of the first building, which we will call \( x \) meters.
Let BE be the height of the second building, which is given as 120 meters.
The horizontal distance between the buildings (AB) is 70 meters.

Draw a line DC parallel to AB, where C is a point on BE.
Then BC = AD = \( x \) meters.
And EC = BE - BC = \( 120 - x \) meters.
Also, DC = AB = 70 meters.
The angle of depression from the top of the second building (E) to the top of the first building (D) is \( 45^\circ \). So, in the right-angled triangle EDC, \( \angle CED = 45^\circ \).
Using the tangent function in \( \triangle EDC \):
\( \tan 45^\circ = \frac{EC}{DC} \)
\( 1 = \frac{120 - x}{70} \)
Multiply both sides by 70.
\( 70 = 120 - x \)
Now, move \( x \) to one side and numbers to the other.
\( x = 120 - 70 \)
\( x = 50 \) m
Therefore, the height of the first building is 50 meters. This problem illustrates how angles of depression help find unknown heights using basic trigonometry.
In simple words: We used the angle of depression and the distance between buildings to find the height of the shorter building. The first building is 50 meters tall.

🎯 Exam Tip: When dealing with two objects of different heights, always draw a horizontal line from the top of the shorter object to form a rectangle and a right triangle, simplifying the problem.

Question 3. From the top of the tower 60 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt{3}\) = 1.732)
Answer: Let BC be the height of the tower, so BC = 60 m.
Let AD be the height of the lamp post, which we will call \( h \) meters.
Let AB be the distance between the lamp post and the tower, which we will call \( x \) meters.

From the diagram, we have: EC = BC - BE. Since AD = BE = \( h \), then EC = \( 60 - h \).
In the right-angled triangle ABC (using the angle of elevation for the bottom of the lamp post):
\( \tan 60^\circ = \frac{BC}{AB} \)
\( \sqrt{3} = \frac{60}{x} \)

\( \implies \) \( x = \frac{60}{\sqrt{3}} \) ....(1)
In the right-angled triangle DEC (using the angle of elevation for the top of the lamp post):
\( \tan 38^\circ = \frac{EC}{DE} \)
Since DE = AB = \( x \), then:
\( 0.7813 = \frac{60 - h}{x} \)

\( \implies \) \( x = \frac{60 - h}{0.7813} \) ....(2)
Now, we equate equations (1) and (2) to solve for \( h \):
\( \frac{60}{\sqrt{3}} = \frac{60 - h}{0.7813} \)
Multiply both sides by \( 0.7813 \) and \( \sqrt{3} \).
\( 60 \times 0.7813 = \sqrt{3} (60 - h) \)
\( 46.878 = 60\sqrt{3} - h\sqrt{3} \)
Now, substitute the value of \( \sqrt{3} = 1.732 \).
\( 46.878 = 60 \times 1.732 - h \times 1.732 \)
\( 46.878 = 103.92 - 1.732h \)
Rearrange the equation to solve for \( h \).
\( 1.732h = 103.92 - 46.878 \)
\( 1.732h = 57.042 \)

\( \implies \) \( h = \frac{57.042}{1.732} \)

\( \implies \) \( h \approx 32.934 \) m
So, the height of the lamp post is approximately 32.93 meters. It is important to remember that angles of depression are measured from the horizontal line of sight down to the object.
In simple words: We used the tower's height and the two angles to find the lamp post's height. The lamp post is about 32.93 meters tall.

🎯 Exam Tip: When using given trigonometric values like tan 38°, use them accurately. Pay close attention to which angle corresponds to which triangle and ensure you're using the correct sides in your tangent ratio.

Question 4. An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are 60° and 30° respectively. Find the distance between the two boats.
Answer: Let A be the position of the aeroplane and B be the point directly below it on the ground. So, the altitude AB = 1800 m.
Let C and D be the positions of the two boats, sailing in the same direction towards the aeroplane.
The angle of depression to the closer boat (C) is \( 60^\circ \), so \( \angle ACB = 60^\circ \).
The angle of depression to the farther boat (D) is \( 30^\circ \), so \( \angle ADB = 30^\circ \).

Let BC = \( y \) and CD = \( x \). The distance between the two boats is \( x \). So, BD = \( x + y \).
In the right-angled triangle ABC:
\( \tan 60^\circ = \frac{AB}{BC} \)
\( \sqrt{3} = \frac{1800}{y} \)

\( \implies \) \( y = \frac{1800}{\sqrt{3}} \) ....(1)
In the right-angled triangle ABD:
\( \tan 30^\circ = \frac{AB}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{1800}{x + y} \)

\( \implies \) \( x + y = 1800\sqrt{3} \) ....(2)
Substitute the value of \( y \) from (1) into (2):
\( x + \frac{1800}{\sqrt{3}} = 1800\sqrt{3} \)
Subtract \( \frac{1800}{\sqrt{3}} \) from both sides.
\( x = 1800\sqrt{3} - \frac{1800}{\sqrt{3}} \)
To combine these, find a common denominator.
\( x = \frac{1800\sqrt{3} \times \sqrt{3} - 1800}{\sqrt{3}} \)
\( x = \frac{1800 \times 3 - 1800}{\sqrt{3}} \)
\( x = \frac{5400 - 1800}{\sqrt{3}} \)
\( x = \frac{3600}{\sqrt{3}} \)
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \).
\( x = \frac{3600\sqrt{3}}{3} \)
\( x = 1200\sqrt{3} \)
Using the given value \( \sqrt{3} = 1.732 \):
\( x = 1200 \times 1.732 \)
\( x = 2078.4 \) m
The distance between the two boats is 2078.4 meters. This calculation shows how two different angles of depression from the same height can help determine a horizontal distance.
In simple words: We used the aeroplane's height and two angles to find how far apart the two boats are. The distance between the boats is 2078.4 meters.

🎯 Exam Tip: When dealing with two objects in the same line of sight, always use two separate triangles to form two equations and then solve them simultaneously for the unknown distance.

Question 5. From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \( \frac{4h}{\sqrt{3}} \) m.
Answer: Let DB be the height of the lighthouse, so DB = \( h \) meters.
Let B be the foot of the lighthouse.
Let A and C be the positions of the two ships on opposite sides of the lighthouse.

The angle of depression to ship A is \( 60^\circ \), so \( \angle DAB = 60^\circ \).
The angle of depression to ship C is \( 30^\circ \), so \( \angle DCB = 30^\circ \).
Let AB = \( x \) and BC = \( y \). The total distance between the ships is \( AC = x + y \).
In the right-angled triangle ABD:
\( \tan 60^\circ = \frac{DB}{AB} \)
\( \sqrt{3} = \frac{h}{x} \)

\( \implies \) \( x = \frac{h}{\sqrt{3}} \) ....(1)
In the right-angled triangle DBC:
\( \tan 30^\circ = \frac{DB}{BC} \)
\( \frac{1}{\sqrt{3}} = \frac{h}{y} \)

\( \implies \) \( y = h\sqrt{3} \) ....(2)
Now, find the total distance between the ships, \( AC = x + y \):
\( AC = \frac{h}{\sqrt{3}} + h\sqrt{3} \)
To add these, we can rationalize the first term or find a common denominator.
\( AC = \frac{h}{\sqrt{3}} + \frac{h\sqrt{3} \times \sqrt{3}}{\sqrt{3}} \)
\( AC = \frac{h}{\sqrt{3}} + \frac{3h}{\sqrt{3}} \)
\( AC = \frac{h + 3h}{\sqrt{3}} \)
\( AC = \frac{4h}{\sqrt{3}} \) m
Hence, it is verified that the distance between the ships is \( \frac{4h}{\sqrt{3}} \) m. This problem highlights how trigonometry can be used to prove relationships between different lengths and angles.
In simple words: We used the height of the lighthouse and the angles to each ship. By adding the distances from the lighthouse to each ship, we showed that the total distance between them is \( \frac{4h}{\sqrt{3}} \) meters.

🎯 Exam Tip: When ships are on opposite sides, the total distance is the sum of their individual distances from the base of the lighthouse. Clearly define variables and apply the tangent function to each relevant triangle.

Question 6. A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is \( 30\sqrt{3} \) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer: Let DB be the total height of the building, so DB = 90 feet.
Let B be the base of the building, and C be the position of the fountain.
The horizontal distance from the building to the fountain (BC) is \( 30\sqrt{3} \) feet.

Initially, the lift is at D (top of the building). The angle of depression to the fountain C is \( 60^\circ \). In \( \triangle DBC \):
\( \tan 60^\circ = \frac{DB}{BC} \)
\( \sqrt{3} = \frac{90}{30\sqrt{3}} \)
\( \sqrt{3} = \frac{3}{\sqrt{3}} \)
\( \sqrt{3} = \sqrt{3} \) (This confirms the initial angle is consistent with the given dimensions.)
Let the speed of the lift be \( x \) feet/minute.
After 2 minutes, the lift has descended a distance of \( 2 \times x \) feet. Let its new position be A.
So, DA = \( 2x \) feet.
The remaining height of the lift from the ground is AB = DB - DA = \( 90 - 2x \) feet.
At position A, the angle of depression to the fountain C is \( 30^\circ \). In \( \triangle ABC \):
\( \tan 30^\circ = \frac{AB}{BC} \)
\( \frac{1}{\sqrt{3}} = \frac{90 - 2x}{30\sqrt{3}} \)
Multiply both sides by \( 30\sqrt{3} \).
\( 30 = 90 - 2x \)
Add \( 2x \) to both sides and subtract 30 from both sides.
\( 2x = 90 - 30 \)
\( 2x = 60 \)

\( \implies \) \( x = 30 \) feet/minute
To convert this speed to feet/second:
Speed = \( \frac{30 \text{ feet}}{1 \text{ minute}} = \frac{30 \text{ feet}}{60 \text{ seconds}} = 0.5 \) feet/second.
The speed of the descending lift is 30 feet per minute, or 0.5 feet per second. This scenario shows how changes in angles of depression can be used to calculate speed or distances traveled over time.
In simple words: The lift starts high, and the angle to the fountain is big. As it goes down, the angle gets smaller. We used these angles and the distance to the fountain to figure out how fast the lift is moving. The lift moves at 30 feet per minute.

🎯 Exam Tip: Break down dynamic problems into snapshots (initial and final positions). Clearly define the distance traveled and remaining height in terms of speed and time, then apply trigonometric ratios to each snapshot.