GSEB Class 10 Maths Solutions Chapter 9 Some Applications of Trigonometry Exercise 9.1

Question 1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.
Answer: To find the height of the pole, we utilize trigonometry. In the right-angled triangle formed by the pole, the ground, and the rope, we apply the sine function. Given the angle is 30 degrees and the hypotenuse (rope length) is 20 meters, we set up the equation \( \sin 30^\circ = \frac{AB}{AC} \), where AB is the pole's height and AC is the rope's length. Since \( \sin 30^\circ \) is \( \frac{1}{2} \), we have \( \frac{1}{2} = \frac{AB}{20} \). Solving this equation for AB, we find \( AB = \frac{20}{2} = 10 \) meters. Thus, the pole's height is 10 meters.
In simple words: We use the sine function for the given right triangle. The height of the pole (AB) is found by multiplying the rope length (20m) by \( \sin 30^\circ \) (which is \( \frac{1}{2} \)). So, the height is 10 meters.

Exam Tip: Always draw a clear diagram and label all known values and the unknown value you need to find. Choose the correct trigonometric ratio (sine, cosine, or tangent) based on the given sides and angles. Ensure the angles are correctly identified.

Question 2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Answer: To find the original height of the tree, we analyze the situation by forming a right-angled triangle. Let the tree's actual height be BD, where AB is the standing portion and AD is the broken portion that bends to touch the ground at point C. Therefore, AD is equal to AC. In the right triangle ABC, the distance from the foot of the tree (B) to the point where the top touches the ground (C) is 8m, and the angle at C is 30°.
First, we apply the cosine function to find AC: \( \cos 30^\circ = \frac{BC}{AC} \). Substituting the given values, \( \frac{\sqrt{3}}{2} = \frac{8}{AC} \), which gives us \( AC = \frac{16}{\sqrt{3}} \) meters.
Next, we use the tangent function to determine AB: \( \tan 30^\circ = \frac{AB}{BC} \). So, \( \frac{1}{\sqrt{3}} = \frac{AB}{8} \), which means \( AB = \frac{8}{\sqrt{3}} \) meters.
The total height of the tree BD is the sum of AB and AC. Adding these values, \( BD = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} \). To rationalize the denominator, we multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \), resulting in \( \frac{24 \sqrt{3}}{3} = 8 \sqrt{3} \) meters. Thus, the tree's total height is \( 8 \sqrt{3} \) meters.
In simple words: The broken tree forms a right triangle. We use cosine to find the length of the fallen part (AC) and tangent to find the standing part (AB). The total height is the sum of these two lengths. First, \( \cos 30^\circ = \frac{8}{AC} \) gives \( AC = \frac{16}{\sqrt{3}} \). Then, \( \tan 30^\circ = \frac{AB}{8} \) gives \( AB = \frac{8}{\sqrt{3}} \). Adding them up, \( AB + AC = \frac{24}{\sqrt{3}} \), which simplifies to \( 8\sqrt{3} \) meters.

Exam Tip: When a tree breaks and bends, remember that the broken upper part becomes the hypotenuse of the right-angled triangle. The total height of the tree is the sum of the standing part and the bent part, ensure you calculate both correctly.

Question 3. A contractor plans to install two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground. For elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Answer: To find the length of each slide, we consider two separate right-angled triangles.
For younger children (below 5 years): In triangle ABC, the height of the slide (AB) is 1.5 meters, and the angle of inclination (at C) is 30°. Using the sine function, \( \sin 30^\circ = \frac{AB}{AC} \). Substituting the values, \( \frac{1}{2} = \frac{1.5}{AC} \), which means \( AC = 1.5 \times 2 = 3 \) meters.
For elder children: In triangle DEF, the height of the slide (DE) is 3 meters, and the angle of inclination (at F) is 60°. Using the sine function, \( \sin 60^\circ = \frac{DE}{DF} \). Substituting the values, \( \frac{\sqrt{3}}{2} = \frac{3}{DF} \), which gives \( DF = \frac{6}{\sqrt{3}} \). Rationalizing the denominator, \( DF = \frac{6 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{6 \sqrt{3}}{3} = 2 \sqrt{3} \). Approximating \( \sqrt{3} \) as 1.732, \( DF = 2 \times 1.732 = 3.464 \) meters.
Thus, the length of the slide for younger children is 3 meters, and for older children, it is approximately 3.46 meters.
In simple words: For younger kids, the slide is 1.5m high with a 30° angle. Using \( \sin 30^\circ = \frac{1.5}{AC} \), the slide length (AC) is 3 meters. For older kids, the slide is 3m high with a 60° angle. Using \( \sin 60^\circ = \frac{3}{DF} \), the slide length (DF) is \( 2\sqrt{3} \) or about 3.46 meters.

Exam Tip: Always distinguish between the two scenarios clearly. Draw separate diagrams or indicate distinct parts for each case. Remember to rationalize the denominator if your answer contains a square root for a final simplified form.

Question 4. The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower is 30°. Find the height of the tower.
Answer: To determine the height of the tower, we set up a right-angled triangle ABC. AB represents the height of the tower, and BC is the distance from the foot of the tower to the observation point, which is 30 meters. The angle of elevation (at C) is 30°. We use the tangent function because it relates the opposite side (AB) to the adjacent side (BC). So, \( \tan 30^\circ = \frac{AB}{BC} \). Substituting the values, \( \frac{1}{\sqrt{3}} = \frac{AB}{30} \). Solving for AB, we get \( AB = \frac{30}{\sqrt{3}} \). To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{3} \), which gives \( AB = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3} \) meters. Using the approximate value of \( \sqrt{3} \) as 1.732, the height is \( 10 \times 1.732 = 17.32 \) meters. Thus, the tower's height is approximately 17.32 meters.
In simple words: We need to find the tower's height. We use \( \tan 30^\circ = \frac{AB}{30} \), where AB is the height and 30m is the ground distance. This gives \( AB = \frac{30}{\sqrt{3}} \), which simplifies to \( 10\sqrt{3} \) or about 17.32 meters.

Exam Tip: For direct questions involving angle of elevation, the tangent ratio is often used when the height and base distance are given. Remember to rationalize the denominator for a clearer and simplified final answer.

Question 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Answer: To determine the length of the kite string, we form a right-angled triangle ABC. AB represents the height of the kite above the ground, which is 60 meters. AC is the length of the string, and C is the point where the string is tied to the ground. The angle of inclination (at C) is 60°. We use the sine function, as it relates the opposite side (AB) to the hypotenuse (AC). So, \( \sin 60^\circ = \frac{AB}{AC} \). Substituting the values, \( \frac{\sqrt{3}}{2} = \frac{60}{AC} \). Solving for AC, we get \( AC = \frac{120}{\sqrt{3}} \). To rationalize the denominator, we multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \), resulting in \( AC = \frac{120 \sqrt{3}}{3} = 40 \sqrt{3} \) meters. Therefore, the string's length is \( 40\sqrt{3} \) meters.
In simple words: We need the string's length. We know the kite's height (60m) and the angle it makes with the ground (60°). Using \( \sin 60^\circ = \frac{60}{AC} \), the string length AC is \( \frac{120}{\sqrt{3}} \), which becomes \( 40\sqrt{3} \) meters after simplifying.

Exam Tip: When dealing with kites or objects attached by a string, the string usually forms the hypotenuse of the right-angled triangle. 'No slack' means the string is perfectly straight, simplifying the geometry.

Question 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer: To find the distance the boy walked, we first adjust the building's effective height. The boy is 1.5m tall, and the building is 30m tall. So, the effective height from the boy's eye level to the top of the building (AB) is \( 30 - 1.5 = 28.5 \) meters. We form two right-angled triangles: ABC (when the angle is 60°) and ABD (when the angle is 30°), where A is the top of the building, B is the point on the building at the boy's eye level, and D and C are the boy's initial and final positions along the ground.
In \( \triangle ABC \), \( \tan 60^\circ = \frac{AB}{BC} \). So, \( \sqrt{3} = \frac{28.5}{BC} \), which gives \( BC = \frac{28.5}{\sqrt{3}} \) meters.
In \( \triangle ABD \), \( \tan 30^\circ = \frac{AB}{BD} \). So, \( \frac{1}{\sqrt{3}} = \frac{28.5}{BD} \), which gives \( BD = 28.5 \sqrt{3} \) meters.
The distance the boy walked (CD) is the difference between BD and BC: \( CD = BD - BC = 28.5 \sqrt{3} - \frac{28.5}{\sqrt{3}} \).
To simplify, we find a common denominator: \( CD = \frac{28.5 \times 3 - 28.5}{\sqrt{3}} = \frac{85.5 - 28.5}{\sqrt{3}} = \frac{57}{\sqrt{3}} \).
Rationalizing the denominator, \( CD = \frac{57 \sqrt{3}}{3} = 19 \sqrt{3} \) meters. Therefore, the boy walked \( 19\sqrt{3} \) meters towards the building.
In simple words: First, subtract the boy's height from the building's height to get the effective height. Then, use tangent for the two angles of elevation (30° and 60°) to find the two distances from the building (BD and BC). The distance the boy walked is the difference between these two distances: \( CD = BD - BC \). This calculation leads to \( 19\sqrt{3} \) meters.

Exam Tip: Remember to subtract the observer's height from the total height of the object if the angle of elevation is measured from their eye level. This creates a smaller right triangle above the eye line, simplifying calculations.

Question 7. From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively, find the height of the tower.
Answer: To determine the height of the transmission tower, we set up two right-angled triangles from the observation point on the ground. Let AB be the distance from the observation point to the base of the building. BC is the height of the building (20m), and CD is the height of the tower (h). So, the total height BD is \( BC + CD = 20 + h \).
First, consider \( \triangle ABC \) for the building's top. The angle of elevation is 45°. So, \( \tan 45^\circ = \frac{BC}{AB} \). Substituting values, \( 1 = \frac{20}{AB} \), which gives \( AB = 20 \) meters.
Next, consider \( \triangle ABD \) for the tower's top. The angle of elevation is 60°. So, \( \tan 60^\circ = \frac{BD}{AB} \). Substituting values, \( \sqrt{3} = \frac{20+h}{20} \). This leads to \( 20 + h = 20\sqrt{3} \).
Solving for h, \( h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \). Using \( \sqrt{3} \approx 1.732 \), we get \( h = 20(1.732 - 1) = 20(0.732) = 14.64 \) meters. Therefore, the transmission tower's height is 14.64 meters.
In simple words: We need the tower's height. We use two right triangles. For the building, \( \tan 45^\circ = \frac{20}{AB} \) shows \( AB = 20 \) meters. For the tower, \( \tan 60^\circ = \frac{20+h}{20} \) helps us find h. Solving for h gives \( 20(\sqrt{3} - 1) \), which is about 14.64 meters.

Exam Tip: For problems involving a tower on a building, draw two triangles sharing a common base (the distance from the observer). One triangle uses the building's height, the other uses the combined height of the building and tower. This dual triangle approach is key.

Question 8. A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground. The angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45° Find the height of the pedestal.
Answer: To find the height of the pedestal, we utilize trigonometry with two angles of elevation from a single point on the ground. Let BC represent the height of the pedestal (h meters) and AB be the height of the statue (1.6 meters). So, the total height AC is \( 1.6 + h \) meters. Let DC be the distance from the observation point on the ground to the base of the pedestal (x meters).
First, consider the angle of elevation to the top of the pedestal, which is 45°. In right \( \triangle BCD \), \( \tan 45^\circ = \frac{BC}{DC} \). Since \( \tan 45^\circ = 1 \), we have \( 1 = \frac{h}{x} \), so \( h = x \) (Equation 1).
Next, consider the angle of elevation to the top of the statue, which is 60°. In right \( \triangle ACD \), \( \tan 60^\circ = \frac{AC}{DC} \). So, \( \sqrt{3} = \frac{1.6 + h}{x} \) (Equation 2).
Substitute \( x=h \) from Equation 1 into Equation 2: \( \sqrt{3} = \frac{1.6 + h}{h} \).
This gives \( \sqrt{3}h = 1.6 + h \). Rearranging the terms, \( \sqrt{3}h - h = 1.6 \), which simplifies to \( h(\sqrt{3} - 1) = 1.6 \).
So, \( h = \frac{1.6}{\sqrt{3} - 1} \). To rationalize the denominator, we multiply by the conjugate \( \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \).
\( h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{1.6(\sqrt{3} + 1)}{3 - 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \) meters. Therefore, the pedestal's height is \( 0.8(\sqrt{3} + 1) \) meters.
In simple words: We have a statue on a pedestal. For the pedestal, \( \tan 45^\circ = \frac{h}{x} \) means \( h=x \). For the total height (statue + pedestal), \( \tan 60^\circ = \frac{1.6+h}{x} \). By substituting \( x=h \) and solving, we get \( h = 0.8(\sqrt{3} + 1) \) meters. This means the pedestal's height is approximately 1.3856 meters.

Exam Tip: For problems with a combined height (like a statue on a pedestal), draw two right triangles using the same base. The height of one triangle is the pedestal, and the height of the other is the combined height. Use the tangent function for both to establish a relationship.

Question 9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. (CBSE 2008, 2009)
Answer: To determine the building's height, we analyze the scenario with two objects (a building and a tower) and two angles of elevation. Let AB be the height of the building (h meters) and CD be the height of the tower (50 meters). Let BC be the horizontal distance between the foot of the building and the foot of the tower (x meters).
First, consider the angle of elevation of the tower's top from the building's foot, which is 60°. In \( \triangle BCD \), \( \tan 60^\circ = \frac{CD}{BC} \). Substituting values, \( \sqrt{3} = \frac{50}{x} \), which gives \( x = \frac{50}{\sqrt{3}} \) (Equation 1).
Next, consider the angle of elevation of the building's top from the tower's foot, which is 30°. In \( \triangle ABC \), \( \tan 30^\circ = \frac{AB}{BC} \). Substituting values, \( \frac{1}{\sqrt{3}} = \frac{h}{x} \), which gives \( x = h\sqrt{3} \) (Equation 2).
Equating the expressions for x from both equations: \( h\sqrt{3} = \frac{50}{\sqrt{3}} \).
Solving for h, we get \( h = \frac{50}{\sqrt{3} \times \sqrt{3}} = \frac{50}{3} \) meters. Converting to a mixed fraction, \( h = 16\frac{2}{3} \) meters. Thus, the building's height is \( 16\frac{2}{3} \) meters.
In simple words: We need to find the building's height. We set up two tangent equations using the distance between the building and tower (x). From the tower's height (50m) and 60° angle, we get \( x = \frac{50}{\sqrt{3}} \). From the building's height (h) and 30° angle, we get \( x = h\sqrt{3} \). Equating these two expressions for x gives \( h = \frac{50}{3} \) meters.

Exam Tip: When dealing with two objects and cross-angles of elevation, draw two separate right triangles that share a common base. Use the tangent function for each, then solve the system of equations for the unknown height, ensuring careful substitution.

Question 10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Answer: To find the height of the poles and the distances from the observation point, we consider two poles of equal height (h) on opposite sides of an 80-meter wide road. Let AB and ED be the poles, and C be a point on the road between them. Let the distance from C to pole ED be x, so \( CD = x \). Then the distance from C to pole AB is \( BC = 80 - x \).
From the point C, the angles of elevation to the tops of the poles are 30° and 60°.
In \( \triangle ABC \) (for the pole with 30° elevation): \( \tan 30^\circ = \frac{AB}{BC} = \frac{h}{80-x} \). This gives \( 80 - x = h\sqrt{3} \) (Equation 1).
In \( \triangle EDC \) (for the pole with 60° elevation): \( \tan 60^\circ = \frac{ED}{CD} = \frac{h}{x} \). This gives \( h = x\sqrt{3} \) (Equation 2).
Substitute the value of h from Equation 2 into Equation 1: \( 80 - x = (x\sqrt{3})\sqrt{3} \).
This simplifies to \( 80 - x = 3x \). Solving for x, \( 4x = 80 \), so \( x = 20 \) meters.
Now, substitute \( x = 20 \) back into Equation 2 to find h: \( h = \sqrt{3} \times 20 = 20\sqrt{3} \) meters.
So, the height of both poles is \( 20\sqrt{3} \) meters. The distances of the point C from the poles are x = 20 meters and \( 80 - x = 80 - 20 = 60 \) meters. Therefore, the height of the poles is \( 20\sqrt{3} \)m, and the distances from the point are 20m and 60m.
In simple words: We have two poles of height 'h' separated by an 80m road. A point on the road has angles of elevation 30° and 60°. Using tangent for both poles, we get two equations relating 'h' and the distances 'x' and '80-x'. Solving these equations, we find x = 20m, and h = \( 20\sqrt{3} \)m. So, pole heights are \( 20\sqrt{3} \)m, and distances are 20m and 60m.

Exam Tip: For problems involving two objects separated by a distance, where an observation point is between them, split the total distance into two parts (x and total-x). Form two right triangles and use the tangent function to relate the heights and distances, then solve simultaneously.

Question 11. A TV tower stands vertically on the banks of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Answer: To find the tower's height and the canal's width, we form two right-angled triangles. Let AB be the height of the TV tower (h meters) and BC be the width of the canal (x meters). Let D be a point 20m away from C, further along the bank. So, the total base distance BD is \( BC + CD = x + 20 \) meters.
First, consider the point directly opposite the tower. In \( \triangle ABC \), the angle of elevation to the tower's top is 60°. So, \( \tan 60^\circ = \frac{AB}{BC} = \frac{h}{x} \). This gives \( h = x\sqrt{3} \) (Equation 1).
Next, consider the point 20m away. In \( \triangle ABD \), the angle of elevation to the tower's top is 30°. So, \( \tan 30^\circ = \frac{AB}{BD} = \frac{h}{x+20} \). This gives \( x + 20 = h\sqrt{3} \) (Equation 2).
Substitute h from Equation 1 into Equation 2: \( x + 20 = \sqrt{3}(x\sqrt{3}) \).
This simplifies to \( x + 20 = 3x \). Solving for x, \( 2x = 20 \), so \( x = 10 \) meters. This is the canal's width.
Substitute \( x = 10 \) back into Equation 1 to find h: \( h = 10\sqrt{3} \) meters. Therefore, the tower's height is \( 10\sqrt{3} \) meters, and the canal's width is 10 meters.
In simple words: We use two triangles for the tower. For the first point, \( \tan 60^\circ = \frac{h}{x} \) means \( h = x\sqrt{3} \). For the second point, \( \tan 30^\circ = \frac{h}{x+20} \) means \( x+20 = h\sqrt{3} \). Solving these equations gives x = 10m (canal width) and h = \( 10\sqrt{3} \)m (tower height).

Exam Tip: For problems involving a tower on a river bank and two observation points, define the river width as 'x' and use the two given angles of elevation. Set up two tangent equations and solve them simultaneously to find both the height and the width of the canal.

Question 12. From the top of 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Answer: To determine the height of the cable tower, we utilize the given angles of elevation and depression from the top of a building. Let DE be the height of the building (7 meters). Let AC be the height of the cable tower. From the top of the building (D), we draw a horizontal line DB to the tower. B is a point on the tower at the same horizontal level as D.
First, consider the angle of depression from D to the foot of the tower (A), which is 45°. In right \( \triangle DEA \), \( \tan 45^\circ = \frac{DE}{AE} \). Since DE is 7m and AE is the distance between the building and tower (x), \( 1 = \frac{7}{x} \), so \( x = 7 \) meters. Thus, \( DB = AE = 7 \) meters.
Next, consider the angle of elevation from D to the top of the tower (C), which is 60°. In right \( \triangle DBC \), \( \tan 60^\circ = \frac{BC}{DB} \). Let BC be h. Substituting values, \( \sqrt{3} = \frac{h}{7} \), which gives \( h = 7\sqrt{3} \) meters.
The total height of the cable tower AC is \( AB + BC \). Since AB is the height of the building segment at the bottom (equal to DE, so 7m), then \( AC = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \) meters. Therefore, the tower's height is \( 7(\sqrt{3} + 1) \) meters.
In simple words: We find the distance (x) between the building and tower using the angle of depression: \( \tan 45^\circ = \frac{7}{x} \) gives x = 7m. Then, we find the part of the tower above the building (h) using the angle of elevation: \( \tan 60^\circ = \frac{h}{7} \) gives \( h = 7\sqrt{3} \)m. The total tower height is \( 7 + 7\sqrt{3} \)m or \( 7(\sqrt{3} + 1) \)m.

Exam Tip: When an observer is on top of a building looking at a tower, draw a horizontal line from the observer's eye level to the tower. This creates two right triangles: one for the angle of depression (to the base) and one for the angle of elevation (to the top), sharing a common horizontal base.

Question 13. As observed from the top of a 75 m high light house from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships.
Answer: To determine the distance between the two ships, we use the angles of depression from the top of a 75-meter high lighthouse. Let AB be the height of the lighthouse (75m). Let C be the position of the ship closer to the lighthouse, and D be the position of the ship farther away, both on the same side. The angles of depression from A are 45° to C and 30° to D. Due to alternate interior angles, \( \angle ACB = 45^\circ \) and \( \angle ADB = 30^\circ \).
First, consider \( \triangle ABC \) for the closer ship. \( \tan 45^\circ = \frac{AB}{BC} \). Substituting values, \( 1 = \frac{75}{BC} \), which gives \( BC = 75 \) meters.
Next, consider \( \triangle ABD \) for the farther ship. \( \tan 30^\circ = \frac{AB}{BD} \). Substituting values, \( \frac{1}{\sqrt{3}} = \frac{75}{BD} \), which gives \( BD = 75\sqrt{3} \) meters.
The distance between the two ships (CD) is the difference between BD and BC: \( CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) \) meters. Therefore, the distance between the two ships is \( 75(\sqrt{3} - 1) \) meters.
In simple words: We use two right triangles for the lighthouse (75m high) and the ships. For the closer ship, \( \tan 45^\circ = \frac{75}{BC} \) means \( BC = 75 \)m. For the farther ship, \( \tan 30^\circ = \frac{75}{BD} \) means \( BD = 75\sqrt{3} \)m. The distance between the ships is the difference \( BD - BC \), which is \( 75(\sqrt{3} - 1) \)m.

Exam Tip: When observing objects from a height (like a lighthouse) with angles of depression, remember that the angle of elevation from the object to the observer is equal to the angle of depression from the observer to the object due to alternate interior angles. Also, use two triangles, one for each object, and typically subtract the shorter base from the longer base to find the distance between the objects.