Exercise 9.1 Practice
Heights and Distances: Fresh Practice Problems
Q1: Circus Artist
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A circus artist is climbing a 50 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 $30^\circ$.
Given: Length of rope (AC) = 50 m, Angle C = $30^\circ$.
To find: Height of pole (AB).
To find: Height of pole (AB).
Using sine ratio: $\sin 30^\circ = \frac{AB}{AC}$
$\frac{1}{2} = \frac{AB}{50} \Rightarrow AB = \frac{50}{2} = 25$
Height of the pole is 25 m.
Q2: Broken Tree
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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 of $30^\circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is 15 m. Find the height of the tree.
Let broken part be $y$ (hypotenuse) and standing part be $x$ (opposite). Total height = $x + y$.
Base = 15 m. Angle = $30^\circ$.
Base = 15 m. Angle = $30^\circ$.
$\tan 30^\circ = \frac{x}{15} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{15} \Rightarrow x = \frac{15}{\sqrt{3}} = 5\sqrt{3}$ m.
$\cos 30^\circ = \frac{15}{y} \Rightarrow \frac{\sqrt{3}}{2} = \frac{15}{y} \Rightarrow y = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
Total Height = $5\sqrt{3} + 10\sqrt{3} = 15\sqrt{3}$ m.
Height of the tree is $15\sqrt{3}$ m.
Q3: Park Slides
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A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 2 m, and is inclined at an angle of $30^\circ$ to the ground, whereas for elder children, she wants to have a steep slide at a height of 4 m, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?
Case 1 (Younger): Height = 2 m, Angle = $30^\circ$. Find Slide Length ($l_1$).
$\sin 30^\circ = \frac{2}{l_1} \Rightarrow \frac{1}{2} = \frac{2}{l_1} \Rightarrow l_1 = 4$ m.
$\sin 30^\circ = \frac{2}{l_1} \Rightarrow \frac{1}{2} = \frac{2}{l_1} \Rightarrow l_1 = 4$ m.
Case 2 (Elder): Height = 4 m, Angle = $60^\circ$. Find Slide Length ($l_2$).
$\sin 60^\circ = \frac{4}{l_2} \Rightarrow \frac{\sqrt{3}}{2} = \frac{4}{l_2} \Rightarrow l_2 = \frac{8}{\sqrt{3}}$ m.
$\sin 60^\circ = \frac{4}{l_2} \Rightarrow \frac{\sqrt{3}}{2} = \frac{4}{l_2} \Rightarrow l_2 = \frac{8}{\sqrt{3}}$ m.
Lengths are 4 m and $8\sqrt{3}/3$ m.
Q4: Tower Elevation
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The angle of elevation of the top of a tower from a point on the ground, which is 50 m away from the foot of the tower, is $60^\circ$. Find the height of the tower.
Base = 50 m, Angle = $60^\circ$. Let Height be $h$.
$\tan 60^\circ = \frac{h}{50}$
$\sqrt{3} = \frac{h}{50} \Rightarrow h = 50\sqrt{3}$ m.
Height is $50\sqrt{3}$ m.
Q5: Flying Kite
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A kite is flying at a height of 90 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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.
Height = 90 m, Angle = $60^\circ$. Find String Length ($L$).
$\sin 60^\circ = \frac{90}{L} \Rightarrow \frac{\sqrt{3}}{2} = \frac{90}{L}$
$L = \frac{180}{\sqrt{3}} = 60\sqrt{3}$ m.
Length of string is $60\sqrt{3}$ m.
Q6: Walking Towards Building
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A 2 m tall boy is standing at some distance from a 32 m tall building. The angle of elevation from his eyes to the top of the building increases from $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.
Eye level height of building = $32 - 2 = 30$ m.
Distance at 60° (closer): $\tan 60 = \frac{30}{x} \Rightarrow x = \frac{30}{\sqrt{3}} = 10\sqrt{3}$.
Distance at 30° (farther): $\tan 30 = \frac{30}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{y} \Rightarrow y = 30\sqrt{3}$.
Distance walked = $y - x = 30\sqrt{3} - 10\sqrt{3} = 20\sqrt{3}$ m.
Distance walked is $20\sqrt{3}$ m.
Q7: Transmission Tower
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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 30 m high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.
Building height ($h_1$) = 30 m. Let tower height = $h$. Distance of point = $x$.
Angle for bottom (building top) = $45^\circ$, Angle for top (tower top) = $60^\circ$.
Angle for bottom (building top) = $45^\circ$, Angle for top (tower top) = $60^\circ$.
$\tan 45^\circ = \frac{30}{x} \Rightarrow 1 = \frac{30}{x} \Rightarrow x = 30$ m.
$\tan 60^\circ = \frac{30 + h}{x} \Rightarrow \sqrt{3} = \frac{30 + h}{30}$.
$30\sqrt{3} = 30 + h \Rightarrow h = 30(\sqrt{3} - 1)$ m.
Height of tower is $30(\sqrt{3} - 1)$ m.
Q8: Statue on Pedestal
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A statue, 2.4 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 $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.
Let pedestal height = $h$, Base = $x$. Statue = 2.4 m.
Pedestal top = $45^\circ$, Statue top = $60^\circ$.
Pedestal top = $45^\circ$, Statue top = $60^\circ$.
$\tan 45^\circ = \frac{h}{x} \Rightarrow x = h$.
$\tan 60^\circ = \frac{h + 2.4}{x} \Rightarrow \sqrt{3} = \frac{h + 2.4}{h}$.
$h\sqrt{3} - h = 2.4 \Rightarrow h(\sqrt{3}-1) = 2.4 \Rightarrow h = \frac{2.4}{\sqrt{3}-1}$.
Rationalizing: $h = \frac{2.4(\sqrt{3}+1)}{2} = 1.2(\sqrt{3}+1)$.
Height of pedestal is $1.2(\sqrt{3} + 1)$ m.
Q9: Building vs Tower
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The angle of elevation of the top of a building from the foot of the tower is $30^\circ$ and the angle of elevation of the top of the tower from the foot of the building is $60^\circ$. If the tower is 60 m high, find the height of the building.
Tower height = 60 m. Let distance between feet = $x$.
Tower elevation = $60^\circ \Rightarrow \tan 60 = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} = 20\sqrt{3}$.
Tower elevation = $60^\circ \Rightarrow \tan 60 = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} = 20\sqrt{3}$.
Building height = $h$. Building elevation = $30^\circ$.
$\tan 30 = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}$.
$\tan 30 = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{20\sqrt{3}}$.
$h = \frac{20\sqrt{3}}{\sqrt{3}} = 20$ m.
Height of the building is 20 m.
Q10: Two Poles
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Two poles of equal heights are standing opposite each other on either side of the road, which is 100 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.
Height = $h$. Total width = 100. Point divides it into $x$ and $100-x$.
Let 60° be at distance $x$. $h = x\sqrt{3}$.
Let 60° be at distance $x$. $h = x\sqrt{3}$.
Let 30° be at distance $100-x$. $h = \frac{100-x}{\sqrt{3}}$.
Equate $h$: $x\sqrt{3} = \frac{100-x}{\sqrt{3}} \Rightarrow 3x = 100 - x \Rightarrow 4x = 100 \Rightarrow x = 25$.
Height $h = 25\sqrt{3}$. Distances are 25 m and 75 m.
Height is $25\sqrt{3}$ m. Distances: 25 m and 75 m.
Q11: Canal Tower
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A TV tower stands vertically on a bank 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 $60^\circ$. From another point 40 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 $30^\circ$. Find the height of the tower and the width of the canal.
Let width = $x$, Height = $h$.
Angle 60°: $h = x\sqrt{3}$.
Angle 60°: $h = x\sqrt{3}$.
Angle 30° (dist $x+40$): $h = \frac{x+40}{\sqrt{3}}$.
Equate $h$: $x\sqrt{3} = \frac{x+40}{\sqrt{3}} \Rightarrow 3x = x + 40 \Rightarrow 2x = 40 \Rightarrow x = 20$.
$h = 20\sqrt{3}$.
Height: $20\sqrt{3}$ m, Width: 20 m.
Q12: Building & Cable Tower
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From the top of a 10 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.
Building = 10 m. Depression 45° means horizontal distance = 10 m (since $\tan 45 = 1$).
Elevation 60° for top part ($h'$): $\tan 60 = h'/10 \Rightarrow h' = 10\sqrt{3}$.
Total tower height = Building height + Top part = $10 + 10\sqrt{3}$.
Height is $10(1 + \sqrt{3})$ m.
Q13: Lighthouse Ships
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As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ and $45^\circ$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Height = 100 m.
Distance to ship 1 (45°): $d_1 = 100 / \tan 45 = 100$ m.
Distance to ship 1 (45°): $d_1 = 100 / \tan 45 = 100$ m.
Distance to ship 2 (30°): $d_2 = 100 / \tan 30 = 100\sqrt{3}$ m.
Distance between ships = $100\sqrt{3} - 100$.
Distance is $100(\sqrt{3} - 1)$ m.
Q14: Balloon
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A 2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 102 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$. Find the distance travelled by the balloon during the interval.
Effective height = $102 - 2 = 100$ m.
Initial Distance ($d_1$ at 60°): $d_1 = 100 / \sqrt{3}$.
Final Distance ($d_2$ at 30°): $d_2 = 100\sqrt{3}$.
Distance travelled = $100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3}$.
Distance is $\frac{200\sqrt{3}}{3}$ m.
Q15: Highway Car Time
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A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^\circ$, which is approaching the foot of the tower with a uniform speed. 10 seconds later, the angle of depression of the car is found to be $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.
Let height = $h$.
Distance at 30° = $h\sqrt{3}$. Distance at 60° = $h/\sqrt{3}$.
Distance at 30° = $h\sqrt{3}$. Distance at 60° = $h/\sqrt{3}$.
Distance covered in 10s = $h\sqrt{3} - h/\sqrt{3} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.
Distance remaining to cover = $h/\sqrt{3}$ (which is exactly half of the distance covered in 10s).
Since speed is uniform, time taken will be half of 10s.
Time taken is 5 seconds.