the angle of elevation of the top of a hill at the foot of a tower is 60 degree and the angle of depression from the top of the tower of the foot of the hill is 30 degree if the tower is 50 find the height of the hill
Answers
Answer:
Let AB is the Tower of height = h = 50 m.
And, let the Height of Hill CD = H m.
Distance between The root of the tower and hill = BC
Now,
In ΔABC
∠C = 30°
TAN(C) = AB/BC
⇒ TAN(30) = 50/BC
⇒ 1/√3 = 50 /BC
⇒ BC = 50√3 m.
Now,
In ΔBCD,
∠B = 60°
Tan(B) = CD/BC
⇒ Tan(60) = H/BC
⇒ BC√3 = H
⇒ H = 50√3*√3 = 150 m.
Explanation:
Maths
The angle of elevation of the top of a hill from the foot of a tower is 60^{o}60
o
and the angle of elevation of the top of the tower from the foot of the hill is 30^{o}30
o
. If the tower is 50 m high, find the height of the hill.
Asked on December 20, 2019 byavatar
Udit Waswani
ANSWER
Let AB be the tower and CD be the hill. Then, \angle ACB = 30^{o}, \angle CAD = 60^{o}∠ACB=30
o
,∠CAD=60
o
and AB=50 \ mAB=50 m.
Let CD= x \ mCD=x m
In right \triangle BAC△BAC, we have,
\cot 30^{o}=\dfrac{AC}{AB}cot30
o
=
AB
AC
\sqrt{3}=\dfrac{AC}{50}
3
=
50
AC
AC=50\sqrt{3} \ mAC=50
3
m
In right \triangle ACD△ACD, we have,
\tan 60^{o}=\dfrac{CD}{AC}tan60
o
=
AC
CD
\sqrt{3}=\dfrac{x}{50\sqrt{3}}
3
=
50
3
x
x=50\times 3=150 \ mx=50×3=150 m
Therefore, the height of the hill is 150 m.