Usage of law of sines

Question

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?

Answer 1

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 $\text{perimeter} = 10 + 2\dfrac{5}{\cos72.5^\circ}\approx43.255$.

Answer 2

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.