Usage of law of sines The vertex angle of an isosceles triangle is 35 degrees. The length of the base is 10 centimeters. How many centimeters are in the perimeter?
I understand the problem as there are two sides with length 10 and one side of unknown length.
I used laws of sines to find the side corresponding to the 35 degree angle.
$$
\frac{\sin35^\circ}{x} = \frac{\sin72.5^\circ}{10}
$$
I get 6 as the length of the side
The answer is 43.3. They get it by dropping an altitude from the vertex to the base and forming congruent right triangles. What am I doing wrong? 
 A: I'm guessing what is intended is that the $35^\circ$ angle is opposite the side of length $10$, and the two equal sides of unknown length meet at that $35^\circ$ angle.
If you drop that perpendicular, you get one of the congruent halves having angles $90^\circ$ and $35^\circ/2=17.5^\circ$.  The third angle would then be $72.5^\circ$.  In that half, the side that meets the $72.5^\circ$ angle has length $5$.  The hypotenuse would then be $5/\cos72.5^\circ$, and the height would be $5\tan72.5^\circ$.
So $\text{perimeter} = 10 + 2\dfrac{5}{\cos72.5^\circ}\approx43.255$.
A: The other answer does not use the law of sines, as your title states.
You're given that the side of $10$ lies opposite the vertex angle of $35^\circ$.
First find the other two angles (the "base angles"). As the triangle is isoceles, they are equal. Denote one base angle as $\theta$.
By the angle sum of the triangle, $2\theta + 35^\circ = 180^\circ$, giving $\theta = 72.5^\circ$.
Now employ the law of sines. Denote one of the sides adjoining the vertex as $x$.
So $\displaystyle \frac{x}{\sin 72.5^\circ} = \frac{10}{\sin 35^\circ}$.
Solving that gives you $x \approx 16.63$cm.
Clearly there are two such sides with length $x$, giving a perimeter of $2x + 10 \approx 2(16.63) + 10 = 43.26$cm. I'm rounding off to $2$ decimal places.
