# 30-60-90 triangle HELP!

### 30-60-90 triangle HELP!

I have a 30-60-90 triangle.
the hypotenuse is z, the leg is z-5, and the base is blank.
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### Re: 30-60-90 triangle HELP!

The question is ambiguous. The leg of length z- 5 is opposite to which angle?

sin(30)= 1/2. If the hypotenuse has length z and the leg opposite the 30 degree angle has length z- 5 then (z- 5)/z= 1/2 so 2(z- 5)= 2z- 10= z. z= 5.

But $$cos(30)= \frac{\sqrt{3}}{2}$$. If it is the leg adjacent to the 30 degree angle that has length 5 then $$(z- 5)/z= \sqrt{3}/2$$ so $$2(z- 5)= 2z- 10= z\sqrt{3}$$. $$(2- \sqrt{3})z= 10$$. $$z= \frac{10}{2- \sqrt{3}}= 20+ 10\sqrt{3}$$.

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