Exercise-9.1, Class 10th, Maths, Chapter 9, NCERT

Q 1.

Question:
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 rope makes an angle 30° with the ground.

Solution:
Let the height of the pole = AB m, and rope = AC = 20 m.
∠ACB = 30°, right Δ ABC.

$$\sin 30\mathrm{°}=\frac{AB}{AC}\Rightarrow \frac{1}{2}=\frac{AB}{20}\Rightarrow AB=10m\mathrm{.}$$

Height of the pole = 10 m.

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Q 2.

Question:
A tree breaks due to storm and the broken part bends so that the top touches the ground making 30° with it. The distance between the foot of the tree and the point where top touches ground = 8 m. Find the height of the tree.

Solution:
Let the tree break at point B, AB = h m, BD = 8 m (distance on ground).
Broken part AD touches ground, ∠ADB = 30°, so AD = hypotenuse = BD / cos 30°.

$$AD=\frac{8}{\frac{\sqrt{3}}{2}}=\frac{16}{\sqrt{3}}m\mathrm{.}$$

Vertical part = AB = AD sin 30° = $\frac{16}{\sqrt{3}}\times \frac{1}{2}=\frac{8}{\sqrt{3}}$

Total height = AB + BD = $\frac{8}{\sqrt{3}}+$ (?? Wait) hold. Actually BD is ground distance, so total height of tree = AB + BC where BC = AD sin 30°, etc. Correcting:
Let unbroken part = BE and broken part = ED touches ground.
If DE makes 30° with ground and BD = 8 m = DE cos 30°.

$$DE=\frac{8}{\cos 30\mathrm{°}}=\frac{8}{\sqrt{3}\mathrm{/}2}=\frac{16}{\sqrt{3}}m\mathrm{.}$$

Vertical BE = DE sin 30° = $\frac{16}{\sqrt{3}}\times \frac{1}{2}=\frac{8}{\sqrt{3}}$

Hence total height = BE + ED sin 30° = $\frac{8}{\sqrt{3}}+\frac{8}{\sqrt{3}}$=$\frac{16}{\sqrt{3}}$=9.24 m (approx).

Height of tree = 9.24 m (approx).

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Q 3.

Question:
Two slides—one for small children (height 1.5 m, 30° incline) and one for elders (height 3 m, 60° incline). Find the length of each slide.

Solution:

(i) For small children:

$$\sin 30\mathrm{°}=\frac{1.5}{L_1}\Rightarrow \frac{1}{2}=\frac{1.5}{L_1}\Rightarrow L_1=3m\mathrm{.}$$

(ii) For elder children:

$$\sin 60\mathrm{°}=\frac{3}{L_2}\Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{L_2}\Rightarrow L_2=\frac{3\times 2}{\sqrt{3}}=\frac{6}{\sqrt{3}}=2\sqrt{3}=3.46m(approx)\mathrm{.}$$

✅ Lengths: 3 m and 3.46 m (approx).

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Q 4.

Question:
Angle of elevation of top of tower = 30°, distance from tower = 30 m. Find height.

Solution:

$$\tan 30\mathrm{°}=\frac{h}{30}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{30}\Rightarrow h=\frac{30}{\sqrt{3}}=10\sqrt{3}=17.32m\mathrm{.}$$

✅ Height of tower = 17.32 m.

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Q 5.

Question:
Kite height = 60 m, inclination of string = 60°. Find string length (no slack).

Solution:

$$\sin 60\mathrm{°}=\frac{60}{l}\Rightarrow \frac{\sqrt{3}}{2}=\frac{60}{l}\Rightarrow l=\frac{60\times 2}{\sqrt{3}}=\frac{120}{\sqrt{3}}=40\sqrt{3}=69.28m\mathrm{.}$$

✅ Length of string = 69.3 m (approx).

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Q 6.

Question:
A 1.5 m tall boy is standing from a 30 m building; angle of elevation from 30° to 60° as he walks towards building. Find distance walked.

Solution:
Let initial distance = x m, final distance = y m.

At 30°:

$$\tan 30\mathrm{°}=\frac{30-1.5}{x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{x}\Rightarrow x=28.5\sqrt{3}=\mathrm{49.35.}$$

At 60°:

$$\tan 60\mathrm{°}=\frac{28.5}{y}\Rightarrow \sqrt{3}=\frac{28.5}{y}\Rightarrow y=\frac{28.5}{\sqrt{3}}=\mathrm{16.46.}$$

Distance walked = x − y = 49.35 − 16.46 = 32.89 m.

He walked ≈ 32.9 m.

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Q 7.

Question:
From a point on ground, angles of elevation of bottom and top of transmission tower fixed atop a 20 m building are 45° and 60° respectively. Find height of tower.

Solution:
Let tower = h m. Distance from building = x m.

From top:

$$\tan 60\mathrm{°}=\frac{20+h}{x}\Rightarrow \sqrt{3}=\frac{20+h}{x}\Rightarrow x=\frac{20+h}{\sqrt{3}}\mathrm{.}$$

From bottom:

$$\tan 45\mathrm{°}=\frac{20}{x}\Rightarrow 1=\frac{20}{x}\Rightarrow x=20.$$

Equate x’s:

$$20=\frac{20+h}{\sqrt{3}}\Rightarrow 20\sqrt{3}=20+h$$

$$\Rightarrow h=20(\sqrt{3}-1)=20(1.732-1)=14.64m\mathrm{.}$$

Height of tower = 14.64 m.

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Q 8.

Question:
A statue 1.6 m tall stands on a pedestal. From a point on ground, angle of elevation of top of statue = 60°, and of top of pedestal = 45°. Find height of pedestal.

Solution:
Let pedestal height = h m, distance from point = x m.

From top of statue:

$$\tan 60\mathrm{°}=\frac{h+1.6}{x}\Rightarrow \sqrt{3}=\frac{h+1.6}{x}\mathrm{.}$$

From top of pedestal:

$$\tan 45\mathrm{°}=\frac{h}{x}\Rightarrow 1=\frac{h}{x}\Rightarrow x=h\mathrm{.}$$

Substitute:

$$\sqrt{3}=\frac{h+1.6}{h}\Rightarrow \sqrt{3}h=h+1.6\Rightarrow h(\sqrt{3}-1)=1.6$$

$$\Rightarrow h=\frac{1.6}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=0.8(\text{√}3+1)=0.8\times 2.732=2.19m\mathrm{.}$$

Height of pedestal = 2.19 m (approx).

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Q 9.

Question:
Angle of elevation of top of building from foot of tower = 30°, and of top of tower from foot of building = 60°. Tower height = 50 m. Find height of building.

Solution:
Let distance between tower and building = x m, height of building = h m.

From foot of tower:

$$\tan 30\mathrm{°}=\frac{h}{x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\Rightarrow x=h\sqrt{3}\mathrm{.}$$

From foot of building:

$$\tan 60\mathrm{°}=\frac{50}{x}\Rightarrow \sqrt{3}=\frac{50}{x}\Rightarrow x=\frac{50}{\sqrt{3}}\mathrm{}$$

Equate:

$$h\sqrt{3}=\frac{50}{\sqrt{3}}\Rightarrow h=\frac{50}{3}=16.67m\mathrm{.}$$

Height of building = 16.67 m.

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Q 10.

Question:
Two equal poles opposite each other across a road 80 m wide. From a point between them, angles of elevation are 60° and 30°. Find height and distances.

Solution:
Let distance from point to first pole = x m, so to other pole = 80 − x m, height = h m.

From 1st pole (60°):

$$\tan 60\mathrm{°}=\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow h=x\text{√}3.$$

From 2nd pole (30°):

$$\tan 30\mathrm{°}=\frac{h}{80-x}\Rightarrow \frac{1}{\text{√}3}=\frac{h}{80-x}\Rightarrow h=\frac{80-x}{\text{√}3}\mathrm{.}$$

Equate:

$$x\text{√}3=\frac{80-x}{\text{√}3}\Rightarrow 3x=80-x\Rightarrow 4x=80\Rightarrow x=20.$$

So $h=20\text{√}3=34.64m\mathrm{}$

Height of each pole = 34.64 m, Distances = 20 m and 60 m.

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Q 11.

Question:
A TV tower stands on one bank of a canal. From a point on the opposite bank, angle of elevation = 60°. From another point 20 m away on same line, angle of elevation = 30°. Find height of tower and width of canal.

Solution:
Let height = h m, width of canal = x m.

From near point (60°):

$$\tan 60\mathrm{°}=\frac{h}{x}\Rightarrow \text{√}3=\frac{h}{x}\Rightarrow h=x\text{√}3.$$

From far point (30°):

$$\tan 30\mathrm{°}=\frac{h}{x+20}\Rightarrow \frac{1}{\text{√}3}=\frac{h}{x+20}\Rightarrow h=\frac{x+20}{\text{√}3}\mathrm{.}$$

Equate:

$$x\text{√}3=\frac{x+20}{\text{√}3}\Rightarrow 3x=x+20\Rightarrow 2x=20\Rightarrow x=10$$

Then $h=10\text{√}3=17.32m\mathrm{}$

Height = 17.32 m, Width = 10 m.

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Q 12.

Question:
From top of 7 m building, angle of elevation of top of tower = 60°, angle of depression of foot = 45°. Find height of tower.

Solution:
Let tower height = h m, distance = x m.

From angle of depression 45°:

$$\tan 45\mathrm{°}=\frac{7}{x}\Rightarrow x=7.$$

From angle of elevation 60°:

$$\tan 60\mathrm{°}=\frac{h-7}{x}\Rightarrow \text{√}3=\frac{h-7}{7}\Rightarrow h-7=7\text{√}3\Rightarrow h=7(1+\text{√}3)=7(2.732)=19.12m\mathrm{.}$$

Height of tower = 19.12 m.

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Q 13.

Question:
From a 75 m high lighthouse, angles of depression of two ships are 30° and 45°. Find distance between ships on same side of lighthouse.

Solution:
Let distances from lighthouse = x₁ (for 45°) and x₂ (for 30°).

$$\tan 45\mathrm{°}=\frac{75}{x_1}\Rightarrow x_1=75.$$

$$\tan 30\mathrm{°}=\frac{75}{x_2}\Rightarrow \frac{1}{\text{√}3}=\frac{75}{x_2}\Rightarrow x_2=75\text{√}3=129.9$$

Distance between ships = 54.9 m.

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Q14 (redo)

Question (restated briefly).
A 1.2 m tall girl watches a balloon flying horizontally at height 88.2 m above the ground. From her eyes the angle of elevation of the balloon is first ${60}^{\circ }$ and later becomes ${30}^{\circ }$. Find the horizontal distance the balloon travelled during this time.

Solution.

Height of balloon above the girl’s eyes

$$H=88.2-1.2=87.0\text{ m}\mathrm{}$$

Let the horizontal distances of the balloon from the girl’s vertical line at the times when the elevation angles are ${60}^{\circ }$and ${30}^{\circ }$ be $x_1$ and $x_2$ respectively.

Using $\tan \theta ={\displaystyle \frac{\text{opposite}}{\text{adjacent}}}$

For $\theta ={60}^{\circ }$ (first position),

$$\tan {60}^{\circ }=\sqrt{3}=\frac{H}{x_1}\Rightarrow x_1=\frac{H}{\sqrt{3}}=\frac{87}{\sqrt{3}}=29\sqrt{3}(\text{m})\mathrm{.}$$

For $\theta ={30}^{\circ }$ (later),

$$\tan {30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{H}{x_2}\Rightarrow x_2=H\sqrt{3}=87\sqrt{3}(\text{m})\mathrm{.}$$

Distance travelled by the balloon $=x_2-x_1$:

$$x_2-x_1=87\sqrt{3}-29\sqrt{3}=58\sqrt{3}\text{ m}\mathrm{.}$$

Numeric approximation ($\sqrt{3}\approx 1.732$):

$$58\sqrt{3}\approx 58\times 1.732=100.46\text{ m (approx).}$$

Answer: $58\sqrt{3}\text{ m}$ $\approx 100.46\text{ m}\mathrm{}$

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Q15 (redo)

Question (restated briefly).
From the top of a tower a man sees a car on the straight highway at an angle of depression ${30}^{\circ }$. The car moves toward the foot of the tower; 6 seconds later the angle of depression becomes ${60}^{\circ }$. How long (from that second observation) will the car take to reach the foot of the tower?

Solution.

Let tower height = $h$. Let horizontal distances from the foot of the tower to the car at the two observations be $x_1$(when angle ${30}^{\circ }$) and $x_2$ (when angle ${60}^{\circ }$).

Angle of depression equals corresponding angle of elevation downward, so:

For ${30}^{\circ }$:

$$\tan {30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{h}{x_1}\Rightarrow x_1=h\sqrt{3}\mathrm{.}$$

For ${60}^{\circ }$:

$$\tan {60}^{\circ }=\sqrt{3}=\frac{h}{x_2}\Rightarrow x_2=\frac{h}{\sqrt{3}}\mathrm{.}$$

Distance the car covers in 6 s:

$$\mathrm{\Delta }x=x_1-x_2=h\sqrt{3}-\frac{h}{\sqrt{3}}=h\text{\hspace{0.17em}⁣}\left(\frac{3-1}{\sqrt{3}}\right)=\frac{2h}{\sqrt{3}}\mathrm{.}$$

So its speed $v={\displaystyle \frac{\mathrm{\Delta }x}{6}}={\displaystyle \frac{2h}{6\sqrt{3}}}={\displaystyle \frac{h}{3\sqrt{3}}}\mathrm{}$

Distance remaining to the foot of the tower at the second observation = $x_2={\displaystyle \frac{h}{\sqrt{3}}}\mathrm{}$

Time to reach foot from that instant:

$$t=\frac{x_2}{v}=\frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}}=3\text{ seconds}\mathrm{.}$$

Answer: The car will reach the foot of the tower in 3 seconds from the second observation.