Two people standing on the same side of a tower in straight line with it, measures the angles of elevation of the height of tower as $25 °$ and $50 °$ respectively. If the height of the tower is 70m, ____ is the distance between the two people.

Question

Accepted Answer

Person 1 is assumed to be standing at point C at a distance of

x_{1}

from point B. Person 2 is assumed to be at point D at a distance of

x_{2}

from point B. The angle of elevation of the top of the tower for person 1 and person 2 are

50 °

and

25 °

respectively.
Now, in the right angle triangle ABC, we have,
AB = h
BC =

x_{1}

∠ACB=

50 °

Using, tan

θ =

(perpendicular / base), where

θ

is the angle of elevation, we get,
⇒

\tan 50 ° = \frac{AB}{BC}

⇒

\tan 50 ° = \frac{h}{x_{1}}

⇒

x_{1} = \frac{h}{\tan 50 °}

- (1)
Now, in right angle triangle ABD, we have,
AB = h
BD =

x_{2}

<ADB=25°
Using, tan

θ =

(perpendicular / base),we get,
⇒ tan

25 ° = \frac{h}{x_{2}} =

\frac{AB}{BD}

⇒

x_{2} = \frac{h}{\tan 25 °}

- (2)
Now, we can clearly see that the total distance between the two people is CD=BD−BC=

x_{2} - x_{1}

.So, subtracting equations (2) and (1), we get,
⇒

x_{2} - x_{1} = \frac{h}{\tan 25 °} - \frac{h}{\tan 50 °}

⇒

x_{2} - x_{1} = h [\frac{1}{\tan 25 °} - \frac{1}{\tan 50 °}]

Using the conversion, tan

θ = \frac{sinθ}{cosθ}

, we get,
⇒

x_{2} - x_{1} = h [\frac{\cos 25 °}{\sin 25 °} - \frac{\cos 50 °}{\sin 50 °}]

Substituting h = 70, we get,
⇒

x_{2} - x_{1} = h [\frac{\sin 25 ° \cos 25 ° - \sin 50 ° \cos 50 °}{\sin 25 ° \sin 50 °}]

Using the identity, sinAcosB−cosAsinB=sin(A−B), we get,
⇒

x_{2} - x_{1} = 70 \times [\frac{\sin (50 ° - 25 °)}{\sin 25 ° \sin 50 °}]

⇒

x_{2} - x_{1} = 70 \times \frac{\sin 25 °}{\sin 25 ° \sin 50 °}

⇒

x_{2} - x_{1} = \frac{70}{\sin 50 °}

Hence, the distance between two people is

\frac{70}{\sin 50 °}

.

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