# find missing angle?

• Jan 24th 2011, 09:22 AM
andyboy179
find missing angle?
hi,
i need to try and find the missing angle from this shape;
Attachment 20577

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

thanks
• Jan 24th 2011, 09:33 AM
e^(i*pi)
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
• Jan 24th 2011, 09:35 AM
Ackbeet
Your approach is certainly correct (law of sines). I get a very slightly different answer:

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

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

$?=\sin^{-1}(0.6\,\sin(96^{\circ}))\approx 36.63^{\circ}.$
• Jan 24th 2011, 09:52 AM
Quote:

Originally Posted by andyboy179
hi,
i need to try and find the missing angle from this shape;
Attachment 20577

would this be correct, sin?= 6xsin96/10= 0.6,

(6sin96)/10=(6/10)sin96=0.6sin96

inverse sin(0.6)= 36.87

I got the impression the answer was prematurely rounded to 0.6 since $\sin(96) \approx 1$