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 $60^\circ$ 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 $\alpha.$
$CD = CE + ED$
$\Rightarrow CD = CE + AB$
$\Rightarrow 30 = CE + 1.5$
$\Rightarrow CE = 30 - 1.5 = 28.5m$
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 $30^\circ$. 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 $60^\circ$ 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 $60^\circ$ 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 $45^\circ$.

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Question 82 Marks

A balloon is connected to a meteorological ground station by a cable of length $215m$ inclined at $60^\circ$ 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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