CBSE Class 10 Mathematics Applications of Trigonometry Assignment Set 11

Short Answer Type Questions

Question 1. In the middle of a 100 m wide river, there is a small island and a tall tree stands on that island. M and N are points directly opposite to each other on the two banks and in line with the tree. If the angles of elevation of the top of the tree from M and N are \( 30^\circ \) and \( 45^\circ \) respectively, find the height of the tree.

Answer:
Let the tree be AB of height h m, with the foot B on the line MN, \( MB = x \) m and \( BN = (100 - x) \) m.
In \( \triangle ABM \): \( \tan 30^\circ = \frac{h}{x} \Rightarrow x = \sqrt{3}h \). In \( \triangle ABN \): \( \tan 45^\circ = \frac{h}{100 - x} \Rightarrow 100 - x = h \).
Adding: \( 100 = \sqrt{3}h + h \Rightarrow h = \frac{100}{\sqrt{3} + 1} = 50(\sqrt{3} - 1) \).
The height of the tree is \( 50(\sqrt{3} - 1) \) m (about 36.6 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 2. An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer. [Exemplar]
Answer:
The top of the tower is \( 22 - 1.5 = 20.5 \) m above the observer's eye level and the horizontal distance is 20.5 m.
\( \tan \theta = \frac{20.5}{20.5} = 1 \Rightarrow \theta = 45^\circ \).
The angle of elevation is \( 45^\circ \).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 3. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are \( 45^\circ \) and \( 60^\circ \) respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use \( \sqrt{3} = 1.73 \)) [CBSE Delhi Set I, II, III, 2016]

Answer:
Let the tower be H m high and the horizontal distance be d m. The top of the building is \( (H - 50) \) m below the top of the tower.
Bottom of the building: \( \tan 60^\circ = \frac{H}{d} \Rightarrow d = \frac{H}{\sqrt{3}} \). Top of the building: \( \tan 45^\circ = \frac{H - 50}{d} \Rightarrow d = H - 50 \).
\( \frac{H}{\sqrt{3}} = H - 50 \Rightarrow H(\sqrt{3} - 1) = 50\sqrt{3} \Rightarrow H = \frac{50\sqrt{3}}{\sqrt{3} - 1} = 25\sqrt{3}(\sqrt{3} + 1) = 75 + 25\sqrt{3} \).
With \( \sqrt{3} = 1.73 \): \( H = 75 + 43.25 = 118.25 \) m and \( d = H - 50 = 68.25 \) m.
The tower is 118.25 m high and the building is 68.25 m away from it.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 4. An airplane while flying at a height of 4000 m from the ground passes vertically above another airplane at an instant when the angles of the elevation of the two planes from the same point on the ground are \( 60^\circ \) and \( 45^\circ \) respectively. Find the vertical distance between the two planes at that instant.
Answer:
Let the point on the ground be at horizontal distance d from the vertical line of the planes. The higher plane is at 4000 m with elevation \( 60^\circ \): \( \tan 60^\circ = \frac{4000}{d} \Rightarrow d = \frac{4000}{\sqrt{3}} \) m.
The lower plane has elevation \( 45^\circ \), so its height is \( d\tan 45^\circ = \frac{4000}{\sqrt{3}} \approx 2309.4 \) m.
Vertical distance \( = 4000 - \frac{4000}{\sqrt{3}} = \frac{4000(3 - \sqrt{3})}{3} \approx \) 1690.6 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Long Answer Type Questions

Question 5. From the top of a tower h metres high, the angles of depression of two objects, which are in line with the foot of the tower are \( \alpha \) and \( \beta \) (\( \beta > \alpha \)). Find the distance between the two objects. [Exemplar]

Answer:
Let AB be the tower, and C, D the two objects with \( \angle ACB = \beta \) and \( \angle ADB = \alpha \) (alternate angles), both on the same side of the tower.
In \( \triangle ABC \): \( \tan \beta = \frac{h}{BC} \Rightarrow BC = h\cot \beta \). In \( \triangle ABD \): \( \tan \alpha = \frac{h}{BD} \Rightarrow BD = h\cot \alpha \).
\( CD = BD - BC = h\cot \alpha - h\cot \beta \).
The distance between the two objects is \( h(\cot \alpha - \cot \beta) \) m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 6. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are \( 45^\circ \) and \( 60^\circ \) respectively. Find the height of the tower. (Use \( \sqrt{3} = 1.73 \)) [CBSE 2016]
Answer:
Let the building be BC (20 m) and the tower AB (h m); D is the point on the ground, \( CD = x \).
\( \tan 45^\circ = \frac{20}{x} \Rightarrow x = 20 \). \( \tan 60^\circ = \frac{20 + h}{x} \Rightarrow 20 + h = 20\sqrt{3} \).
\( h = 20(\sqrt{3} - 1) = 20 \times 0.73 = 14.6 \).
The height of the tower is 14.6 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 7. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be \( 60^\circ \) and \( 30^\circ \) respectively. Find the height of the balloon above the ground. [Exemplar]
Answer:
The lower window is 2 m and the upper window is \( 2 + 4 = 6 \) m above the ground. Let the balloon be H m above the ground at horizontal distance d. From the lower window the angle is \( 60^\circ \), from the upper window \( 30^\circ \).
\( H - 2 = d\tan 60^\circ = \sqrt{3}d \) and \( H - 6 = d\tan 30^\circ = \frac{d}{\sqrt{3}} \).
Subtracting: \( 4 = d\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = \frac{2d}{\sqrt{3}} \Rightarrow d = 2\sqrt{3} \) m.
\( H = 2 + \sqrt{3} \times 2\sqrt{3} = 2 + 6 = 8 \).
The height of the balloon above the ground is 8 m.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 8. At the foot of a cliff the elevation of its summit is \( 45^\circ \). After ascending 1000 m towards the cliff up a slope of inclination \( 30^\circ \), the elevation of the summit is found to be \( 60^\circ \). Find the height of the cliff.
Answer:
Let the summit be H m above the starting level, so its horizontal distance from the starting point is \( H\cot 45^\circ = H \) m.
After ascending 1000 m up the \( 30^\circ \) slope, the climber has risen \( 1000\sin 30^\circ = 500 \) m and moved horizontally \( 1000\cos 30^\circ = 500\sqrt{3} \) m towards the cliff.
From there the summit is \( (H - 500) \) m higher and \( (H - 500\sqrt{3}) \) m away horizontally, with elevation \( 60^\circ \): \( H - 500 = \sqrt{3}(H - 500\sqrt{3}) = \sqrt{3}H - 1500 \).
\( H(\sqrt{3} - 1) = 1000 \Rightarrow H = \frac{1000}{\sqrt{3} - 1} = 500(\sqrt{3} + 1) \).
The height of the cliff is \( 500(\sqrt{3} + 1) \) m (about 1366 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 9. A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is \( 45^\circ \). The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is \( 30^\circ \). Find the speed of flying of the bird.
Answer:
Let the point be O. The bird is at P, 80 m above the ground, with \( \angle POA = 45^\circ \), so \( OA = 80 \) m. After 2 seconds it is at D with elevation \( 30^\circ \), so \( OA' = \frac{80}{\tan 30^\circ} = 80\sqrt{3} \) m.
Distance flown \( PD = 80\sqrt{3} - 80 = 80(\sqrt{3} - 1) = 80 \times 0.732 = 58.56 \) m.
Speed \( = \frac{58.56}{2} = 29.28 \) m/s.
The speed of the bird is 29.28 m/s.

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.

Question 10. Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of \( 30^\circ \). Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be \( 45^\circ \). Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak. [CBSE OD Set-I, 2019]

Answer:
Height of the bird above the ground \( PQ = 200\sin 30^\circ = 100 \) m.
The bird is \( 100 - 50 = 50 \) m above Deepak's level. If the distance of the bird from Deepak is x m: \( \sin 45^\circ = \frac{50}{x} \Rightarrow x = 50\sqrt{2} \).
The distance of the bird from Deepak is \( 50\sqrt{2} \) m (about 70.7 m).

Teacher's Note:
1. Draw a neat figure and mark the right angle and the given angles first.
2. Choose the ratio (tan, sin or cos) that connects the given side and the required side.