3 Marks Question

Question 13 Marks

The angle of depression from the top of a tower of a point $A$ on the ground is $30^\circ .$ On moving a distance of $20$ metres from the point $A$ towards the foot of the tower to a point $B,$ the angle of elevation of the top of the tower from the point $B$ is $60^\circ $. Find the height of the tower and its distance from the point $A.$

Answer

Let $PQ$ be the tower of height, $h$ metres.
Let $A$ be the first point and $B$ be the point after moving a distance of $20m$.
In right $\triangle\text{APQ},$
$\cot30^\circ=\frac{\text{AP}}{\text{PQ}}$
$\Rightarrow\sqrt{3}=\frac{\text{x}+20}{\text{h}}$
$\Rightarrow\text{h}=\frac{\text{x}+20}{\sqrt{3}}\dots(\text{i})$
In right $\triangle\text{BPQ},$
$\cot60^\circ=\frac{\text{BP}}{\text{PQ}}$
$\Rightarrow\frac{1}{\sqrt{3}}=\frac{\text{x}}{\text{h}}$
$\Rightarrow\text{h}=\text{x}\sqrt{3}\dots(\text{ii})$
From $(i)$ and $(ii),$
$\frac{\text{x}+20}{\sqrt{3}}=\text{x}\sqrt{3}$
$\Rightarrow\text{x}+20=3\text{x}$
$\Rightarrow2\text{x}=20$
$\Rightarrow\text{x}=10$
From $(ii),$ we have
So, $\text{h}=10\sqrt{3}=10\times1.732=17.32\text{m}$
Distance of the tower from point $A = (x + 20)m = 30m$
Hence, the height of the tower is $17.32m$
And the distance of the tower from point $A$ is $30m.$

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Question 23 Marks

From the top of a tower $100m$ high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression $30^\circ $ and $45^\circ $ respectively. Find the distance between the cars. $\big[\text{Take}\sqrt{3}=1.732\big]$

Answer

Let the distance between the cars $= AC$
Height of the tower $= 100m$
In right $\triangle\text{ABD},$
$\tan45^\circ=\frac{\text{BD}}{\text{AB}}$
$\Rightarrow1=\frac{100}{\text{AB}}$
$\Rightarrow\text{AB}=100\text{m}$
In right $\triangle\text{CBD},$
$\frac{\text{BD}}{\text{BC}}=\tan30^\circ$
$\Rightarrow\frac{100}{\text{BC}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\text{BC}=100\sqrt{3}\text{m}$
$\text{AC}=\text{AB}+\text{BC}$
$=100+100\sqrt{3}$
$=100(1+\sqrt{3})\text{m}$
$=273.2\text{m}$
$\therefore$ The distance between the cars is $273.2m.$

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Question 33 Marks

The angles of elevation of the top of a tower from two points at distances of $4\ m$ and $9\ m$ from the base of the tower and in the same straight line with it are complementary. Show that the height of the tower is $6$ metres.

Answer

Let one angle of elevation be $\theta.$
Since the angle are comlementary, the other angle is $(90^\circ-\theta).$
From right $\triangle\text{CAB},$ we have
$\frac{\text{AB}}{\text{AC}}=\tan\theta$
$\Rightarrow\frac{\text{h}}{4}=\tan\theta$
$\Rightarrow\text{h}=4\tan\theta\dots(\text{i})$
From right $\triangle\text{CAB},$ we have
$\frac{\text{AB}}{\text{AD}}=\tan(90^\circ-\theta)$
$\Rightarrow\frac{\text{h}}{9}=\tan(90^\circ-\theta)$
$\Rightarrow\text{h}=9\tan(90^\circ-\theta)$
Multiplying $(i)$ and $(ii)$ we get
$\text{h}^2=36\tan\theta\times\tan(90^\circ-\theta)$
$\Rightarrow\text{h}^2=36\tan\theta\times\cot\theta$
$\Rightarrow\text{h}^2=36$
$\Rightarrow\text{h}=6\text{m}$
Thus, the height of the tower is $6\ m.$
Hence proved.

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