A surveyor wants to find out the height of a tower. He measures ∠A as tan A= 3/4 . What
is the height of the tower if A is 40 m from its base as shown in the figure?
Answers
Answer:
The height of the tower is 22.5 meters.
Step-by-step explanation:
A surveyor measure the angle at A and finds that :
\tan A=\dfrac{3}{4}tanA=
4
3
\begin{gathered}A=\tan^{-1}(\dfrac{3}{4})\\\\A=36.86^{\circ}\end{gathered}
A=tan
−1
(
4
3
)
A=36.86
∘
It i given that, A is 30 m from its base. We know that,
\tan A=\dfrac{P}{B}tanA=
B
P
, P is perpendicular distance and B is base
Here, P is the height of the tower
\begin{gathered}P=B\times \tan A\\\\P=30\times \dfrac{3}{4}\\\\P=22.5\ m\end{gathered}
P=B×tanA
P=30×
4
3
P=22.5 m
So, the height of the tower is 22.5 meters.
The height of the tower is 22.5 meters.
Step-by-step explanation:
A surveyor measure the angle at A and finds that :
\tan A=\dfrac{3}{4}tanA=
4
3
\begin{gathered}A=\tan^{-1}(\dfrac{3}{4})\\\\A=36.86^{\circ}\end{gathered}
A=tan
−1
(
4
3
)
A=36.86
∘
It i given that, A is 30 m from its base. We know that,
\tan A=\dfrac{P}{B}tanA=
B
P
, P is perpendicular distance and B is base
Here, P is the height of the tower
\begin{gathered}P=B\times \tan A\\\\P=30\times \dfrac{3}{4}\\\\P=22.5\ m\end{gathered}
P=B×tanA
P=30×
4
3
P=22.5 m
So, the height of the tower is 22.5 meters.
Answer:
Explanation: