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ICSE Class X Math 2014 Paper

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Question : 52 of 52
Marks: +1, -0
An aeroplane at an altitude of 250 m250\text{ m} observes the angle of depression of two boats on the opposite banks of a river to be 45∘45^{\circ} and 60∘60^{\circ} respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution:  
Let AD=250 mAD = 250\text{ m} height of aeroplane
Two boats are at B and C.
Let BD=xBD = x and DC=yDC = y
From △ADB\triangle ADB;
 From △ADB;x250=cot⁡45∘\text{ From } \triangle ADB ; \frac{x}{250} = \cot 45^{\circ}
x250=1\frac{x}{250} = 1
⇒x=250 m\Rightarrow x = 250\text{ m}
From △ADC;y250=cot⁡60∘\triangle ADC ; \frac{y}{250} = \cot 60^{\circ}
y250=13\frac{y}{250} = \frac{1}{\sqrt{3}}
⇒y=250×13 m\Rightarrow y = 250 \times \frac{1}{\sqrt{3}}\text{ m}
Width of river BC=BD+DCBC = BD + DC
=x+y= x + y
=250+2503= 250 + \frac{250}{\sqrt{3}}
=250(1+13)= 250\left(1 + \frac{1}{\sqrt{3}}\right)
=250(3+13)= 250\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)
=250(1.732+11.732)=250(2.7321.732)= 250\left(\frac{1.732+1}{1.732}\right) = 250\left(\frac{2.732}{1.732}\right)
=250×1.577= 250 \times 1.577
=394.25 m=394 m.= 394.25\text{ m} = 394\text{ m}.
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