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°. 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°. 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° and 45° 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 4m and 9m 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 6m.
Hence proved.

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