Question

A statue is mounted on top of a 21 foot hill. From the base of the hill to where you are standing is 57feet and the statue subtends an angle of 7.1° to where you are standing. Find the height of the statue.

Answer 1

Answer:1.1405 foot

Step-by-step explanation:

Let height of statue be h and angle subtended by base of statue is $\theta$

from diagram

$tan\theta =\frac{21}{57}$

and $tan\left ( \theta +7.1\right )=\frac{21+h}{57}$

and we know $tan\left ( A+B\right )=\frac{tanA+tanB}{1-tanAtanB}$

using above formula

$\frac{tan\theta +tan7.1}{1-tan\theta tan7.1}=\frac{21+h}{57}$

$57\left ( 0.3684+tan7.1\right )=\left ( 21+h\right )\left ( 1-0.3684\times tan7.1\right )$

$21+tan7.1=\left ( 21+h\right )\left ( 0.95411\right )$

h=1.1405 foot

Answer 2

Please find the attached diagram for a better understanding of the question.

As we can see from the diagram,

RQ = 21 feet = height of the hill

PQ = 57 feet = Distance between you and the base of the hill

SR= h=height of the statue

$\angle SPR=7.1^0$=Angle subtended by the statue to where you are standing.

$\angle x$=$\angleRPQ$ which is unknown.

Let us begin solving now. The first step is to find the angle $\angle x$ which can be found by using the following trigonometric ratio in $\Delta PQR$:

$tan(x)=\frac{RQ}{PQ}=\frac{21}{57}$

Which gives x to be:

$x=tan^{-1} (\frac{21}{57})\approx 20.22^{0}$

Now, we know that $\angle x$ and $\angle SPR$ will get added to give us the complete angle $\angle SPQ$ in the right triangle $\Delta PQS$.

We can again use the tan trigonometric ratio in $\Delta PQS$ to solve for the height of the statue, h.

This can be done as:

$tan(\angle SPQ)=\frac{SQ}{PQ}$

$tan(7.1^0+20.22^0)=\frac{SR+RQ}{PQ}$

$tan(27.32^0)=\frac{h+21}{57}$

$\therefore h+21=57\times tan(27.32^0)$

$h\approx8.45 feet$

Thus, the height of the statue is approximately, 8.45 feet.