[3 marks sum]

Question 13 Marks

Calculate $BC $.

Answer

In $\triangle ADC ,$
$\frac{C D}{A D}=\tan 42^{\circ}$
$\Rightarrow C D=20 \times 0.9004=18.008 m $
$\operatorname{In} \triangle ADB$
$\frac{A D}{B D}=\tan 35^{\circ}$
$\Rightarrow B D=\frac{A D}{\tan 35^{\circ}}=\frac{20}{0.7002}=28.563 m$
$\therefore BC = BD - CD =10.55 m $

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Question 23 Marks

In the following diagram , $AB$ is a floor $-$ board : $\text{PQRS}$ is a cubical box with each edge $= 1m$ and $\angle B = 60^\circ $. Calculate the lenght of the board $AB.$

Answer

In $\triangle PSB$
$\frac{P S}{P B}=\sin 60^{\circ}$
$\Rightarrow P B=\frac{2}{\sqrt{3}}=1.155 m $
In $\triangle APQ$
$\angle APQ = 60^\circ$
$\therefore \frac{P Q}{A P}=\cos 60^{\circ}$
$\Rightarrow A P=\frac{1}{\frac{1}{2}}=2 m$
$\therefore A B=A P+P B=2+1.155=3.155 m$

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Question 33 Marks

The angle of elevation of the top of a tower is observed to be$ 60^\circ$ . At a point, $30 m$ vertically above the first point of observation, the elevation is found to be $45^\circ$ . Find:
$(i)$ the height of the tower,
$(ii)$ its horizontal distance from the points of observation.

Answer

In $\triangle ABC$
$\tan 60^{\circ}=\frac{ h +30}{ x }$
$x \sqrt{3}= h +30 \ldots . .(1)$
$\ln \triangle ADE $
$\tan 45=\frac{ h }{ x }$
$1=\frac{ h }{ x }$
$h = x \ldots . . .(2)$
From equation $(1)$
$x \sqrt{3}=x+30$
$x(\sqrt{3}-1)=30$
$x=\frac{30}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$x=\frac{30(\sqrt{3}+1)}{3-1}=\frac{30(1.732+1)}{2}$
$=15 \times 2.732$
$=40.980 m $
From equation $(2)$
$h = 40.98 m$
Height of tower $= h + 30 = 40.98 + 30$
$= 70.98 m$
$= 71 m (approx.)$
Horizontal distance $= x = 40.98 m = 41 m.$

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Question 43 Marks

From the figure , given below , calculate the lenth of CD .

Answer

InΔ AED
$\frac{ AE }{ DE }=\tan 22^{\circ}$
⇒ AE = DE tan 22°= 15 x 0.404=6.06 m
InΔ ABC,
$\frac{ AB }{ BC }=\tan 47^{\circ}$
⇒ AB = BC tan 47° = 15 x 1.072= 16.09 m
∴ CD = BE = AB- AE= 10.03 m

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Question 53 Marks

In the figure, given below, it is given that $AB$ is perpandiculer to $BD$ and is of length $X$ metres. $DC = 30m .\angle ADB = 30^\circ$ and $\angle ACB = 45^\circ . $Without using tables, find $X$

Answer

In $\triangle ABC$
$\frac{A B}{B C}=\tan 45^{\circ}=1$
$\Rightarrow B C=A B=X$
$\ln \triangle ABD ,$
$\frac{A B}{B D}-\tan 30^{\circ}$
$\Rightarrow \frac{x}{30+x}-\frac{1}{\sqrt{3}}$
$\Rightarrow 30=(\sqrt{3}-1) x$
$\therefore x=\frac{30}{1.732-1}=40.98 m$

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Question 63 Marks

A kite is attached to a string. Find the length of the string, when the height of the kite is $60 m$ and the string makes an angle $30$ with the ground

Answer

Let the lenght of the rope be $x \ m$
$\sin 30^{\circ}=\frac{60}{x}$
$\Rightarrow \frac{1}{2}=\frac{60}{x}$
$\therefore x=120 m$
so the lenght of the rope is $120 m$

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Question 73 Marks

A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is $2.4 \ m$ from the wall and the ladder is making an angle of $680$ with the ground. Find the height, upto which the ladder reaches.

Answer

Let the height upto which the ladder reaches be $h \ m$ .
Given that angle of elevation is $68^\circ$
$\tan 68^{\circ}=\frac{h}{2.4}$
$\Rightarrow 2.475=\frac{h}{2.4}$
$\therefore h=2.475 \times 2.4$
$=5.94 m$
so, the ladder reaches upto a height of $5. 94\ m$

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Question 83 Marks

The angle of elevation of the top of a tower from a point on the ground and at a distance of $160\ m$ from its foot, is found to be $60^\circ$. Find the height of the tower.

Answer

Let the height of the tower be $h \ m.$
Given that angle of elevation is $60^\circ$
$\tan 60^{\circ}=\frac{h}{160}$
$\Rightarrow \sqrt{3}=\frac{h}{160}$
$\therefore h=160 \sqrt{3}=277.12 m$
so height of the tower is $277.12. m$

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Question 93 Marks

From the top of a cliff $92\ m$ high, the angle of depression of a buoy is $20^\circ$. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.a

Answer

Let $AB$ be the cliff and $C$ be the buoy.
Given $, AB = 92\ m$
Also $,\angle ACB= 20^\circ$
$\therefore \frac{ AB }{ BC }=\tan 20^{\circ}$
$\therefore BC =\frac{92}{0.3640}$
$=252.7\ m \approx 253\ m $
Hence, the buoy is at a distance of $253\ m$ from the foot of the cliff.

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Mathematics [3 marks sum]

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