Find height of the triangle (antenna) given angle of elevation and displacement from base > A man standing near a radio station antenna observes that the angle of elevation to the top of the antenna is $A = 63°$. He then walks $s = 130$ feet further away and observes that the angle of elevation to the top of the antenna is $B = 49°$ (see the figure below). Find the height of the antenna to the nearest foot. (Hint: Find $x$ first.) > > ![]( I get $362\ \mathrm{ft}$, but apparently that's incorrect.

$$\frac{\text{height}}{\text{distance}}=\tan(63°)\quad\text{and}\quad\frac{\text{height}}{\text{distance}+130}=\tan(49°)$$

therefore

$$\begin{align} &&\tan(63°)\times d&=d\times\tan(49°)+130\times\tan(49°)\\\ \implies &&d&=\frac{130\times\tan(49°)}{(\tan(63°)-\tan(49°))}\\\ \implies &&\text{height}&=\frac{\tan(63°)\times130\times\tan(49°)}{(\tan(63°)-\tan(49°))} \end{align}$$

And that, my friend, is what you have ($\approx361.35$).