Question

the angle of elevation from a point on the ground to the top of a pyramid is 21°40. the angle of elevation from a point 125 feet farther back to the top of the pyramid is 10°20. find the height of the pyramid. what is the height of the pyramid? ft (round to the nearest integer)

Explanation:

Step1: Set up tangent - ratio equations

Let the height of the pyramid be $h$ and the distance from the closer point to the base of the pyramid be $x$. From the closer - point angle of elevation ($21^{\circ}40'$), we know that $\tan(21^{\circ}40')=\frac{h}{x}$ since $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ in a right - triangle formed by the height of the pyramid, the distance from the point on the ground to the base of the pyramid, and the line of sight to the top of the pyramid. From the farther - point angle of elevation ($10^{\circ}20'$), we know that $\tan(10^{\circ}20')=\frac{h}{x + 125}$.
First, convert the angles from degrees and minutes to decimal degrees. $21^{\circ}40'=21+\frac{40}{60}\approx21.67^{\circ}$ and $10^{\circ}20'=10+\frac{20}{60}\approx10.33^{\circ}$.
So we have $\tan(21.67^{\circ})=\frac{h}{x}$, which gives $x=\frac{h}{\tan(21.67^{\circ})}$, and $\tan(10.33^{\circ})=\frac{h}{x + 125}$, which gives $x + 125=\frac{h}{\tan(10.33^{\circ})}$.

Step2: Substitute $x$ into the second equation

Substitute $x=\frac{h}{\tan(21.67^{\circ})}$ into $x + 125=\frac{h}{\tan(10.33^{\circ})}$:
$\frac{h}{\tan(21.67^{\circ})}+125=\frac{h}{\tan(10.33^{\circ})}$.
We know that $\tan(21.67^{\circ})\approx0.400$ and $\tan(10.33^{\circ})\approx0.183$.
The equation becomes $\frac{h}{0.400}+125=\frac{h}{0.183}$.
Multiply through by $0.400\times0.183$ to clear the fractions:
$0.183h+125\times0.400\times0.183 = 0.400h$.
$0.183h+9.15 = 0.400h$.

Step3: Solve for $h$

Subtract $0.183h$ from both sides:
$9.15=0.400h - 0.183h$.
$9.15 = 0.217h$.
Then $h=\frac{9.15}{0.217}\approx42$.

Answer:

$42$