2 Marks Questions

Question 12 Marks

In Fig., what are the angles of depression from the observing position $O_1$ and $O_2$ of the object at $A$?

Answer

In the given figure, draw $XY || ABC$ from $O_1, O_2$ (by joinging them)
$\because\ \angle\text{AO}_1\text{C}=60^\circ$
$\angle\text{AO}_1\text{O}_2=90^\circ-60^\circ=30^\circ$
Similarly,
$\because\ \angle\text{O}_2\text{AB}=45^\circ$
$\angle\text{XO}_2\text{A}=\angle\text{O}_2\text{AB}=45^\circ$ (alternate angles)
Hence angles of depression are $30^\circ$ and $45^\circ$.

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Question 22 Marks

B is a pole of height 6m standing at a point B and CD is a ladder inclined at angle of 60° to the horizontal and reaches upto a point D of pole. If AD = 2.54m, find the length of the ladder. $(\text{Use }\sqrt{3}=1.73)$

Answer

BD = AB - AD = 6 - 2.54 = 3.46m

In rt., $\triangle\text{DBC,}$
$\sin60^\circ=\frac{\text{BD}}{\text{DC}}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\frac{3.46}{\text{DC}}$
$\Rightarrow\ \sqrt{3}\text{DC}=3.46\times2$
$\Rightarrow\ \text{DC}=\frac{3.46\times2}{1.73}=4\text{m}$
Length of the ladder, DC = 4

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Question 32 Marks

An observer, 1.5m tall, is 28.5m away from a 30m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer

Let AB = 1.5m be the observer and CD = 30m be the tower.
Let the angle of elevation of the top of the tower be $\alpha.$
CD = CE + ED
⇒ CD = CE + AB
⇒ 30 = CE + 1.5
⇒ CE = 30 - 1.5 = 28.5m
In $\triangle\text{CEB,}$
$\tan\alpha=\frac{\text{CE}}{\text{BE}}=\frac{28.5}{28.5}$
$\Rightarrow\ \tan\alpha=1$
$\Rightarrow\ \tan\alpha=\tan45^\circ$
$\Rightarrow\ \alpha=45^\circ$

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Question 42 Marks

An observer, 1.7m tall, is $20\sqrt{3}\text{m}$ away from a tower. The angle of elevation from the eye of an observer to the top of tower is 30°. Find the height of the tower.

Answer

$\text{CB}=\text{DE}=20\sqrt{3}\text{m}$

In $\triangle\text{ABC,}$
$\frac{\text{AB}}{\text{BC}}=\tan30^\circ$
$\frac{\text{H}-1.7}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$
$(\text{H}-1.7)\sqrt{3}=20\sqrt{3}$
$\text{H}-1.7=20$
$\text{H}=20+1.7=21.7\text{m}$

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Question 52 Marks

What is the angle of elevation of the Sun when the length of the shadow of a vetical pole is equal to its height?

Answer

Let height of a vertical pole = h
Then its shadow = x

$\therefore$ h = x
Let $\theta$ be the angle of elevation of the sun
Then, $\tan\theta=\frac{\text{AB}}{\text{BC}}=\frac{\text{h}}{\text{x}}=\frac{\text{h}}{\text{h}}$
$\Rightarrow\ \tan\theta=1=\tan45^\circ$ $(\because\ \tan45^\circ=1)$
$\Rightarrow\ \theta=45^\circ$

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Question 62 Marks

A ladder 15m long just reaches the top of a vertical wall. If the ladders makes an angle of 60° with the wall, then find the height of the wall.

Answer

Given that, the height of the ladder = 15m
Let the height of the vertical wall = h
and the ladder makes an angle of elevation 60° with the wall i.e., $\theta=60^\circ.$

In $\triangle\text{QPR},\ \cos60^\circ=\frac{\text{PR}}{\text{PQ}}=\frac{\text{h}}{15}$
$\Rightarrow\ \frac{1}{2}=\frac{\text{h}}{15}$
$\Rightarrow\ \text{h}=\frac{15}{2}\text{m}=7.5\text{m}$
Hence, the required heigth of the wall 7.5m.

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Question 72 Marks

Find the angle of elevation of the sum (sun's altitude) when the length of the shadow of a vertical pole is equal to its height.

Answer

Let $\theta$ be the angle of elevation of sun. Let AB be the vertical pole of height h and BC be the shadow of equal length h.
Here we have to find angle of elevation of sun.
We have the corresponding figure as follows,

So we use trigonometric ratios to find the required angle.
In a triangle ABC,
$\Rightarrow\ \tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan\theta=\frac{\text{h}}{\text{h}}$
$\Rightarrow\ \tan\theta=1$
$\Rightarrow\ \theta=45^\circ$
Hence the angle of elevation of sun is 45°.

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Question 82 Marks

A balloon is connected to a meteorological ground station by a cable of length 215m inclined at 60° to the horizontal. Determine the height of the balloon from the ground. Assume that there is no slack in the cable.

Answer

Length of cable connected to balloon [CB] = 215m
Angle of inclination of cable with ground $\alpha=60^\circ$
Height of balloon from ground = 'h'm = AB
The above data is represented in form of figure as shown
In right triangle one of the included angle is $\theta$ then
$\sin\theta=\frac{\text{Opposite side}}{\text{hypotenuse}}$
$\sin60^\circ=\frac{\text{AB}}{\text{BC}}\Rightarrow\ \frac{\sqrt{3}}{2}=\frac{\text{h}}{215}$
$\Rightarrow\ \text{h}=\frac{215\sqrt{3}}{2}=107.5\sqrt{3}\text{m}$
$\therefore$ Height of balloon from ground $=107.5\sqrt{3}\text{m}$

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Maths 2 Marks Questions

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