# Thread: [SOLVED] Law of Sines

1. ## [SOLVED] Law of Sines

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.]

Thanks.

Garrett

2. 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.
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: .$\displaystyle \text{Perimeter} \:=\:60 + 60 + 81.1 \:=\:201.1\text{ feet}$

3. Thank you Soroban.

I had drawn an incorrect diagram.

Garrett

Originally Posted by garrett88
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.]

Thanks.

Garrett