Question

Ab is a straight road leading at C the foot of a tower. A being at distance of 200 m from C and B is 125 m nearer. If the angle of elevation of top of the tower at B is double the angle of elevation at A find the height of the tower?

Answer 1

Answer:

The height of the tower is 100m.

Step-by-step explanation:

According to Question AB is a straight road leading at C which is the foot of tower DC.

Let height of tower be 'h' m.

Given: AC=200 m, AB=75 m, BC=125 m, ∠DBC=2Ф and ∠DAC=Ф

Formula :-

1. $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$

2. $\tan(2\theta)=\dfrac{2 tan \theta}{1+tan^{2}\theta}$

In ΔADC,  By formula 1

$\tan\theta =\dfrac{DC}{AC}$

$\tan\theta =\dfrac{h}{200}$    ........equation-(1)

In ΔBDC,  By formula 1

$\tan2\theta =\dfrac{DC}{BC}$

$\tan2\theta =\dfrac{h}{125}$    .........equation-(2)

$\dfrac{2\tan\theta}{1+\tan^{2}\theta } =\frac{h}{125}$

Putting value of $\tan\theta =\dfrac{h}{200}$ from  equation-(1) in equation-(2)

$\dfrac{2(h/200)}{1+(h/200)^{2} }=\dfrac{h}{125}$

$\dfrac{h/100}{1+(h/200)^{2} }=\dfrac{h}{125}$

$\dfrac{h/100}{1+h^{2}/40000 }=\dfrac{h}{125}$

$\dfrac{h/100}{(40000+h^{2})/40000 }=\dfrac{h}{125}$

$\dfrac{40000h}{100(40000+h^{2}) }=\dfrac{h}{125}$

$\dfrac{400}{40000+h^{2} }=\dfrac{1}{125}$

$50000=40000+h^{2}$

$h^{2}=50000-40000$

$h^{2}=10000$

$h=\sqrt{10000}$

h=100m

Hence, The height of tower is 100 m