30-60-90 triangle HELP!

30-60-90 triangle HELP!

Postby Guest » Sat Jan 25, 2020 4:21 pm

I have a 30-60-90 triangle.
the hypotenuse is z, the leg is z-5, and the base is blank.
I'm supposed to find the value of z, but I'm stumped. Please help!

Re: 30-60-90 triangle HELP!

Postby HallsofIvy » Sun Jan 26, 2020 8:09 am

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

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