How do you find the base of an isosceles triangle when given the legs only?

Question

This is an SAT math problem, and it's really confusing me.

https://i.stack.imgur.com/aZJNl.png

I am completely lost. Do I have to use trigonometry? I know that the angles of the triangle are 30, 30, and 120, but I do not know where to go after that.

Answer 1

If you cut the bottom triangle in half by bisecting the bottom line from the top angle, you split the $120^\circ$ angle in half to give you $60^\circ$. You know the other angle of the bottom triangle is $30^\circ$. You've reduced the problem to a $30^\circ - 60^\circ-90^\circ$ triangle relation. So the side lengths for a $30^\circ - 60^\circ-90^\circ$ triangle are $x: 30^\circ$, $2x: 90^\circ$, and $x\sqrt3: 60^\circ$.

Since the hypotenuse of this new triangle is $6$ and opposite the $90^\circ$, then $x=3$. Then the bottom line of the new triangle is $3\sqrt3$. Since we bisected this triangle, this is only half the required length. So we double it to get $6\sqrt3$.

Answer 2

SO $\;\angle AGF=120^\circ\;$ , so the basis angles are each $\;30^\circ\;$ , so you can use the sines theorem:

$$\frac{AG}{\sin 30^\circ}=\frac{AF}{\sin120^\circ}\implies AF=\sqrt3\cdot AG$$

Answer 3

All you really need to know to solve this is that in a $30°-60°-90°$ triangle the hypotenuse is twice as long as the short leg, and the long leg is $\sqrt{3}$ times as long as the short leg.

In this particular case, draw a dotted line from $G$ to $AF$ dividing it into two $30°-60°-90°$ triangles. Then since $AG$, the hypotenuse, is $6$ units long, you can easily find the length of the dotted line (which would be the short leg), then use that to find the length of $AF$.

In general, the SAT does not expect you to know any trigonometry other than the "special triangles" ($30°-60°-90°$ and $45°-45°-90°$). Those show up a lot, and you should be on the lookout for them.