1 Marks Question

Question 11 Mark

The ratio of the height of a tower and the length of its shadow on the ground is $\sqrt{3}: 1$. What is the angle of elevation of the Sun?

Answer

Let the angle of elevation of the Sun be $\theta$.
Let the height of the tower be $p$.
And the length of the shadow be $b$.

It is given than ratio of height to the length of the shadow is $=\sqrt{3}: 1$
$\Rightarrow p: b=\sqrt{3}: 1$
We know that $\tan \theta$ is given by perpendicular over base
$\Rightarrow \tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{A B}{B C}$
$\Rightarrow \tan \theta=\frac{p}{b}$
$\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}=\sqrt{3}$
$\therefore \theta=60^{\circ}\left(\because \tan 60^{\circ}=\sqrt{3}\right)$
Hence, the angle of elevation of the sun is $60^{\circ}$.

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Question 21 Mark

If a tower $30 m$ high, easts a shadow $10 \sqrt{3} m$ long on the ground, then what is the angle of elevation of the sun?

Answer

Given:
$AB=30 m($ Height $)$
$BC=10 \sqrt{3} m ($ Length $)$
Let the angle of elevation be $\theta$
In $\triangle \text{ABC}$
$\tan \theta=\frac{AB}{BC}$
$\Rightarrow \tan \theta=\frac{30}{10 \sqrt{3}}$
$\Rightarrow \tan \theta=\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{3 \sqrt{3}}{3}$
$\Rightarrow \tan \theta=\sqrt{3}$
As $\tan 60^{\circ}=\sqrt{3}$
$\theta=60^{\circ}$
Therefore, the angle of elevation of the sun is $60^{\circ}$

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Question 31 Mark

A ladder leaning against a wall, makes an angle of $60^{\circ}$ with the horizontal. If the foot of the ladder is $2.5 m$ away from the wall, find the length of the ladder.

Answer

In the given figure $\text{AB}$ is ladder,

$\frac{\text { Base }}{\text { Hypotenuse }}=\cos 60^{\circ}$
Using $\cos 60^{\circ}=\frac{1}{2}$ we get,
$\frac{B C}{A B}=\frac{1}{2}$
$A B=2 B C$
$A B=2 \times 2.5$
$A B=5 m$
Therefore, the length of the ladder is $5 m .$

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Question 41 Mark

In Fig. $1, \text{AB}$ is a 6 m high pole and $\text{CD}$ is a ladder inclined at an angle of $60^{\circ}$ to the horizontal and reaches up to a point $D$ of pole. If $\text{AD}=2.54 m$, find the length of the ladder. $($Use $\sqrt{3}=1.732 )$

Answer

Given: $\text{AB}=6 m, \text{AD}=2.54 m$ and $\text{CD}$ is inclined at an angle of $60^{\circ}$
To find: Length of the ladder i.e. $\text{CD}$
Solution: From the figure, we can see that,
$\text{AB=AD+DB}=6 m$
Since, $\text{AD}=2.54 m$
So, $2.54 m+\text{DB}=6 m$
$\text{DB}=3.46 m$
Now in the $\triangle \text{BCD}$,
$\frac{BD}{CD}=\sin 60^{\circ}$
$\Rightarrow \frac{3.46}{CD}=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{3.46}{C D}=\frac{1.732}{2}$
$\Rightarrow \text{CD}=\frac{2 \times 3.46}{1.732}$
$\Rightarrow \text{CD}=3.99 m$
Hence, the length of the ladder $\text{CD}$ is $3.99 m \approx 4 m$.

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Question 51 Mark

In figure, a tower $A B$ is $20 m$ high and $B C$, its shadow on the ground, is $20 \sqrt{3} m$ long. Find the Sun's altitude.

Answer

$A B=p=20 m \text { (Given) }$
$B C=b=20 \sqrt{3} m \text { (Given) }$
Now,
$\tan \theta=\frac{p}{b}$
$\tan \theta=\frac{20}{20 \sqrt{3}}=\frac{1}{\sqrt{3}}$
$\tan \theta=\tan 60^{\circ}$
$\theta=60^{\circ}$

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Question 61 Mark

The tops of two towers of height $x$ and $y$, standing on level ground, subtend angles of $30^{\circ}$ and $60^{\circ}$ respectively at the centre of the line joining their feet, then find $x: y$.

Answer

Let $T$ is the centre of the line joining the feet of the two towers $\text{PR}$.
$\operatorname{In} \triangle \text{QPT}$
$\tan 30^{\circ}=\frac{QP}{PT}$
Using $\tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { we get, }$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{P T}$
$\Rightarrow \text{PT}=\sqrt{3} x$
Also,
In $\triangle \text{SRT}$
$\tan 60^{\circ}=\frac{RS}{TR}$
$\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow \sqrt{3}=\frac{y}{TR}$
$\Rightarrow \text{TR}=\frac{y}{\sqrt{3}}$
Since it is given that $T$ is the centre of $\text{PR}$ this implies that $\text{PT=TR}$
$\Rightarrow \sqrt{3} x=\frac{y}{\sqrt{3}} \Rightarrow \frac{x}{y}=\frac{1}{3}$
Hence, the ratio of $x$ and $y$ is $1: 3$.

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Question 71 Mark

In figure, the angles of depressions from the observing positions $O _1$ and $O _2$ respectively of the object A are ________________ .

Answer

From $O _1$, Angle of depression is
$
90^{\circ}-60^{\circ}=30^{\circ}
$
From $O _2$ = Angle of depression is
$
90^{\circ}-45^{\circ}=45^{\circ}
$
Ans. $3 0 ^{\circ}, 4 5 ^{\circ}$

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Question 81 Mark

In figure$-2,$ the angle of elevation of the top of a tower $AC$ from a point $B$ on the ground is $60^{\circ}$. If the height of the tower is $20 m ,$ find the distance of the point from the foot of the tower.

Answer

In $\triangle ABC , \tan 60^{\circ}=\frac{ AC }{ AB }$
$\Rightarrow \sqrt{3}=\frac{20}{ AB }$
$\Rightarrow AB =\frac{20}{\sqrt{3}} m$
$\Rightarrow AB =\frac{20 \sqrt{3}}{3} m$

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