Hello, Garrett!

Need help with the following problem involving the Law of Sines:

A corner of a park occupies a triangular area that faces two streets

that meet at an angle of 85 degrees.

The sides of the area facing the street are each 60 feet in length.

A landscaper wants to plant flowers around the edges of the triangular area.

Find the perimeter of the triangular area.

[The answer is 201.1 feet.] Code:

A
*
* *
* 85° *
60 * * 60
* *
* *
* *
* 47.5° 47.5° *
B * * * * * * * * * C
x

If we __must__ use the Law of Sines, here we go . . .

Triangle ABC is isoceles: $\displaystyle AB = AC$

. . Hence: .$\displaystyle \angle B = \angle C$

Since: .$\displaystyle A + B + C \:=\:180^o$, we have: .$\displaystyle \angle B \,= \,\angle C \,= \,47.5^o$

Law of Sines: .$\displaystyle \frac{x}{\sin85^o} \:=\:\frac{60}{\sin47.5^o} \quad\Rightarrow\quad x \:=\:\frac{60\sin85^o}{\sin47.5^o} \:=\:81.07082...$

. . Hence: .$\displaystyle x\:\approx\:81.1$ feet.

Therefore: .$\displaystyle \text{Perimeter} \:=\:60 + 60 + 81.1 \:=\:201.1\text{ feet}$