[5 marks sum]

Question 15 Marks

There is a building of height 7 m next to a cable tower of unknown height. From the top of the building, the angle of elevation of the top of the tower is $60^{\circ}$ and the angle of depression to the foot of the tower is $45^{\circ}$. Find the height of the cable tower.

Answer

Let AB be the cable tower and CD be the building.
Let $AB =h m$ and $BD =x$
$\begin{array}{l} CD =7 m \text { and } AE =y m \\ \therefore EB = CD =7 m\end{array}$
Now in $\triangle AEB , \tan 60^{\circ}=\frac{ AE }{ CE }$
$\Rightarrow \sqrt{3}=\frac{y}{x} \Rightarrow y=\sqrt{3} x\quad \ldots(i) $
$\text { In } \triangle CDB, \tan 45^{\circ}=\frac{CD}{BD} \Rightarrow 1=\frac{7}{x} \Rightarrow x=7 m\quad \ldots(ii) $
From (i) and (ii), we have
$y=\sqrt{3} \times 7 \Rightarrow y=7 \sqrt{3} m$
$\therefore$ Height of cable tower $= BE + AE =7+7 \sqrt{3}=7(1+\sqrt{3}) m$.

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Question 25 Marks

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the plane to two points on both banks of a river in opposite directions are $45^{\circ}$ and $60^{\circ}$ respectively. Find the width of the river (Use $\sqrt{3}=1.73$ )

Answer

Let A be the position of aeroplane. C and D are the two points on bank of river in opposite sides.
$\begin{array}{l}\text {Let } BD =x \text { and } BC =y \\ \angle ADB =60^{\circ} \text { and } \angle ACB =45^{\circ}\end{array}$
Now in $\triangle ABC , \tan 45^{\circ}=\frac{ AB }{ BC } \Rightarrow 1=\frac{300}{y} \Rightarrow y=300 m$
In $\triangle ABD , \tan 60^{\circ}=\frac{ AB }{ BD } \Rightarrow \sqrt{3}=\frac{300}{x} \Rightarrow x=\frac{300}{\sqrt{3}}$
Width of river $= BC + BD =300+\frac{300}{\sqrt{3}}=\frac{300 \sqrt{3}+300}{\sqrt{3}}$
$=\frac{300(\sqrt{3}+1)}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=100(3+\sqrt{3}) m$

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Question 35 Marks

Two poles of equal heights are standing opposite to each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of poles are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of poles and the distances of the point from the poles.

Answer

Let AB and CD are two poles of equal height $h$.
$\tan 60^{\circ}=\frac{AB}{BE} \Rightarrow \sqrt{3}=\frac{h}{x} \Rightarrow h=\sqrt{3} x\quad \ldots(i) $
In $\triangle C D E, \tan 30^{\circ}=\frac{C D}{D E} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{80-x}$
From (i), $\frac{1}{\sqrt{3}}=\frac{\sqrt{3} x}{80-x} \Rightarrow 3 x=80-x$
$\begin{array}{l}\Rightarrow x=20 m \Rightarrow BE =20 m \\ \text {From (i), } h=\sqrt{3} x=20 \sqrt{3}\end{array}$
$\therefore$ Height of each pole $=20 \sqrt{3} m$.
Also, the point is 20 m from first pole and $(80-20) m =60 m$ from the second pole.

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Question 45 Marks

Two towers AB and CD are standing at some distance apart. From the top of tower AB , the angle of depression of the foot of tower CD is $30^{\circ}$. From the top of tower CD , the angle of depression of the foot of tower AB is $60^{\circ}$. If the height of tower CD is ' $h$ ' m , then prove that the height of tower AB is $\frac{h}{3} \mathrm{~m}$.

Answer

In $\triangle ABD$,
$\tan 30^{\circ}=\frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{x} \Rightarrow x=\sqrt{3} AB\quad \ldots(i) $
In $\triangle CDB , \tan 60^{\circ}=\frac{ CD }{ BD } \Rightarrow \sqrt{3}=\frac{h}{x}\quad \ldots(ii) $
From (i) and (ii), we have
$\sqrt{3}=\frac{h}{\sqrt{3} AB } \Rightarrow 3 AB =h \Rightarrow AB =\frac{h}{3} m$
$\therefore$ Height of tower $AB =\frac{h}{3} m$. Proved.

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Question 55 Marks

An observer measures angles of elevation of two towers of equal height from a point between the towers. The angles of elevation of the tops of the two towers from this point are $60^{\circ}$ and $30^{\circ}$. If this point is at a distance of 120 m from the first tower, find the distance between the towers.

Answer

Let AB and CD are two towers of equal height $h\ m$.
E is the point between them.
$BE=120\ m \text { and } DE=x\ m$
Now in $\triangle ABE$
$\tan 60^{\circ}=\frac{AB}{BE} \Rightarrow \sqrt{3}=\frac{h}{120}$
$\Rightarrow h=120 \sqrt{3} m$
In $\triangle C D E$,
$\tan 30^{\circ}=\frac{ CD }{ DE } \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{120 \sqrt{3}}{x} \Rightarrow x=360 m$
The distance between two towers $= DE + BE =360 m+120 m=480 m$.

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Mathematics [5 marks sum]

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