Hello, magnaye!

6. From the top of a building 60 ft. high the angle of elevation

of the top of a communications tower is $\displaystyle 13.8^o.$

From the bottom of the building the angle of elevation

of the top of the tower is $\displaystyle 35.2^o.$

Find the Height of the tower.

Also, determine the distance from the building to the tower Code:

* C
* * |
* * |
* * |
* * |
* 13.8 * |
A * - - - - - * - - - - - | y
| * |
| * |
60 | * |
| * |
| * 35.2 |
B * - - - - - - - - - - - * D
x

Let $\displaystyle x \,=\,BD,\;y \,=\,CD$

Since $\displaystyle \angle CBD = 35.2^o$, then: $\displaystyle \angle ABC = 54.8^o$

$\displaystyle \angle CAB \:=\:13.8^o + 90^o \:=\:103.8^o$

Hence: .$\displaystyle \angle ACB \:=\:180^o - 54.8^o - 103.8^o \:=\:21.4^o$

In $\displaystyle \Delta ABC$, Law of Sines: .$\displaystyle \dfrac{BC}{\sin103.8^o} \:=\:\dfrac{60}{\sin21.4^o}$

. . $\displaystyle BC \:=\:\dfrac{60\sin103.8^o}{\sin21.4^o} \:=\: 159.6924203 \;\approx\;159.69$

In right triangle $\displaystyle CDB\!:\;\;\sin35.2^o \:=\:\dfrac{y}{BC}

$

. . $\displaystyle y \:=\:159.69\sin35.2^o \:=\:92.05047657$

Therefore: .$\displaystyle \boxed{y \;\approx\;92.05\text{ ft}}$

In right triangle $\displaystyle CDB\!:\;\;\cos38.2^o \:=\:\dfrac{x}{BC}$

. . $\displaystyle x \:=\:159.69\cos35.2^o \:=\:130.4898688$

Therefore: .$\displaystyle \boxed{x \;\approx\;130.49\text{ ft}}$