Trigonometry Problem

[mattgad]

Triangle ABC is such that BC = 5 root 2cm, angle ABC = 30 degrees, and angle BAC = theta, where sin theta = root 5 over 8. Find AC

My working:

I have called AC, x.

\(\displaystyle \Large{x \over {\sin 30}} = {{5\sqrt 2 } \over {{{\sqrt 5 } \over 8}}}\)

\(\displaystyle \Large{x \over {\sin 30}} = {{40\sqrt 2 } \over {\sqrt 5 }}\)

\(\displaystyle \Large x = \sin 30{{40\sqrt 2 } \over {\sqrt 5 }}\)

\(\displaystyle \Large x = {{20\sqrt 2 } \over {\sqrt 5 }}\)

\(\displaystyle \Large x = 2\sqrt 2\)

The books answer is 4 root 10, where have I gone wrong?

Thanks.

 

[Unco]

Hey, Matt.

\(\displaystyle \mbox{ x = \frac{20\sqrt{2}}{\sqrt{5}} }\) is correct.

Rationalise the denominator to retry simplifying.