[4 marks sum]

Question 14 Marks

The radius of a circle is given as $15 \ cm$ and chord $AB$ subtends an angle of $131^\circ$ at the centre $C$ of the circle. Using trigonometry, calculate:
$(i)$ the length of $AB$;
$(ii)$ the distance of $AB$ from the centre $C$.

Answer

Given $, CA = CB =15 \ cm\angle ACB = 131^\circ$
Drop a perpendicular $CP$ from center $C$ to the
Chord $AB.$
Then $CP$ bisects $\angle ACB$ as well as chord $AB$ .
$\therefore \angle A C P=65.5^{\circ}$
$\text { In } \triangle ACP ,$
$\frac{A P}{A C}=\sin \left(65.5^{\circ}\right)$
$\Rightarrow AP =15 x\ \ x \ \ 0.91=13.65 \ cm$
$\text { (1) } AB =2 AP =2 x\ \ x \ \ 13.65=27.30 \ cm .$
$\text { (2) } CP = AP \cos \left(65.5^{\circ}\right)$
$=15 \times 0.415=6.22 \ cm $

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Question 24 Marks

Answer

In $\triangle AMO$
$\operatorname{COS} 30^{\circ}=\frac{A O}{M O}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{A O}{6}$
$\Rightarrow A O=5.20\ m $
In $\triangle BNO$
$\sin 47^\circ = \frac{O B}{N O}$
$\Rightarrow 0.73=\frac{O B}{5}$
$\Rightarrow O B=3.6\ m$
$A B=O A+O B$
$\Rightarrow A B=5,20+3.65$
$\Rightarrow A B=8.85 \ m$

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Question 34 Marks

A vertical tower is $20\ m$ high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is $0.53$. How far is he standing from the foot of the tower?

Answer

Let $AB$ be the tower of height $20\ m$.
Let $\theta$ be the angle of elevation of the top of the tower from point $C$.
Given $, \cos\theta = 0.53$
$\Rightarrow \theta=58^{\circ}$
$\ln \triangle ABC$
$\frac{A B}{B C}=\tan 58^{\circ}$
$\Rightarrow \frac{20}{B C}=1.6$
$\therefore B C=\frac{20}{1.6}=12.5 \ m $

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Question 44 Marks

From a window $A , 10\ m $ above the ground the angle of elevation of the top $C$ of a tower is $X ^\circ ,$ Where $ \tan x^{\circ}=\frac{5}{2}$ and the angle of depression of the foot $D$ of the tower is $Y^\circ ,$ Where $\tan y^{\circ}=\frac{1}{4}$.calculate the height $CD$ of the tower in metres .

Answer

Here $,AB = DE = 10\ m$
In $\triangle ADE$
$\frac{D E}{A E}=\tan y^{\circ}=\frac{1}{4}$
$\Rightarrow AE = 4DE $
$= 4 x 10 = 40\ m$
In$ \triangle AEC,$
$\frac{C E}{A E}=\tan x^{\circ}=\frac{5}{2}$
$\Rightarrow C E=40 \times \frac{5}{2}=110 \ m$
$C D=D E+E D=(10+100)=110 \ m$
Hence, the height of the tower $CD$ is $110 \ m$.

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Question 54 Marks

The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45°. Find the height of the first tower.

Answer

Let AB and CD be the two towrs.
The height of the first tower is AB= 160 m
The horizontel distance between the tow towers is BD = 75 m
And the angle of depression of the first tower as seen from the top of the second tower is∠ ACE = 45°
InΔ ACE
$\frac{A E}{E C}=\tan 45^{\circ}=1$
⇒ AE = EC =BD = 75 m
∴ CD = EB = AB - AE = (160 - 75 ) = 85 m
Hence , height of the other tower is 85 m .

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Question 64 Marks

The angle of elevation of the top of an unfinished tower at a point distance $80 \ m$ from its base is $30^\circ $. How much higher must the tower be raised so that its angle of elevation at the same point may be $60^\circ$ ?

Answer

Let $AB$ be the unfinished tower and $C$ be the top of the tower when finished .
let $P$ be a point $80\ m $ from the foot $A$ in $\triangle BAP$
$\tan 30^{\circ}=\frac{A B}{A P}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{80}$
$\Rightarrow A B=\frac{80}{\sqrt{3}}=46.19\ m $
in $ \triangle CAP$
$\tan 60^{\circ}=\frac{A C}{A P}$
$\Rightarrow \sqrt{3}=\frac{A C}{80}$
$\Rightarrow AC =80 \sqrt{3}=138.56\ m $
Therefore, the tower must be raised by $(138.56 - 46.19) m= 92.37\ m$

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Question 74 Marks

The upper part of a tree, broken over by the wind, makes an angle of $45^\circ$ with the ground and the distance from the root to the point where the top of the tree touches the ground is $15\ m$. What was the height of the tree before it was broken?

Answer

let the height of the tree after bracking be $h\ m$
here $\theta = 45^\circ$
$\therefore \tan 45^{\circ}=\frac{h}{15}$
$\Rightarrow 1=\frac{h}{15}$
$\therefore h=15 m$
now , lenth of the tree broken by the wind
$=\frac{15}{\sin 45^{\circ}}$
$=15 \sqrt{2}$
$=21.21 m$
so, height of the tree before it was broken is $(15 + 21.21 )m = 36.21 \ m.$

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Question 84 Marks

A man stands 9 m away from a flag $-$ pole. He observes that angle of elevation of the top of the pole is $28^\circ$ and the angle of depression of the bottom of the pole is $13^\circ$. Calculate the height of the pole.

Answer

Let $AB$ be the man and $PQ$ be the flag $-$ pole
given $, AR = 9\ m.$
Also $,\angle PAR = 28^\circ$ and $\angle QAR = 13^\circ$
$\therefore \frac{P R}{A R}=\tan 28^{\circ}$
$\Rightarrow P R=9 \times 0.532=4.788\ m$
Also $, \frac{R Q}{A R}=\tan 13^{\circ}$
$\Rightarrow R Q=9 \times 0.231=2.079\ m$
Hence , height of the pole $= PR + RQ = 6.867\ m$

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Question 94 Marks

Two climbers are at points $A$ and $B$ on a vertical cliff face. To an observer $C, 40\ m$ from the foot of the cliff, on the level ground $, A $ is at an elevation of $48^\circ$ and $B$ of $57^\circ$ . What is the distance between the climbers?

Answer

let $P$ be the foot of the cliff on level ground
then $, \angle ACP = 48 ^\circ$ and $\angle BCP = 57 ^\circ$
$\therefore \frac{B P}{P C}=\tan 57^{\circ}$
$\angle BP =40 \times 1.539=61.57\ m$
Aslo $, \frac{A P}{P C}=\tan 48^{\circ}$
$\Rightarrow AP = 40 x \ 1.110 = 44. 4\ m$
Hence, distance between the climbers $= AB = BP - AP = 17.17\ m$

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Question 104 Marks

Two vertical poles are on either side of a road. $A 30 \ m$ long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle $32^\circ 24\ '$ with the pole and when it is turned to rest against another pole, it makes angle $32^\circ 24\ '$ with the road. Calculate the width of the road.

Answer

Let $AB$ the ladder and $\angle ABP = 32^\circ 24 .$
$\therefore \frac{B P}{A B}=\sin 32^{\circ} 24$
$\Rightarrow B P=30 \times 0.536=16.08 .$
When rotated, let the ladder be $AC$ and $\angle CAQ= 32^\circ 24$.
$\frac{B Q}{B C}=\cos 32^{\circ} 24$
$\Rightarrow B Q=30 \times 0.844=25.32$
Hence, width of the road $= (16.08 + 25.32 )= 41.4\ m$

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Mathematics [4 marks sum]

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