find missing angle?

**andyboy179** · January 24th 2011, 08:22 AM

hi,
i need to try and find the missing angle from this shape;

would this be correct, sin?= 6xsin96/10= 0.6, inverse sin(0.6)= 36.87
answer= 36.87????

thanks

**e^(i*pi)** · January 24th 2011, 08:33 AM

It's better to avoid using x to mean multiply. 6sin(96) is easy to understand. Let $\theta$ be the angle you need to find

$\sin \theta = \dfrac{6\sin(96)}{10}$

Never round off halfway through a question, especially to one decimal place!

$\theta = \arcsin \left( \dfrac{6\sin(96)}{10}\right) = 36.6^{\circ} \text{ 3sf }$

You've got the method, you just need to look at your rounding

**Ackbeet** · January 24th 2011, 08:35 AM

Your approach is certainly correct (law of sines). I get a very slightly different answer:

$\dfrac{\sin(?)}{6}=\dfrac{\sin(96^{\circ})}{10}$

$\sin(?)=0.6\,\sin(96^{\circ})$

$?=\sin^{-1}(0.6\,\sin(96^{\circ}))\approx 36.63^{\circ}.$