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

answer= 36.87????

thanks

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- Jan 24th 2011, 08:22 AMandyboy179find 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

answer= 36.87????

thanks - Jan 24th 2011, 08:33 AMe^(i*pi)
It's better to avoid using x to mean multiply. 6sin(96) is easy to understand. Let $\displaystyle \theta$ be the angle you need to find

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

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

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

You've got the method, you just need to look at your rounding - Jan 24th 2011, 08:35 AMAckbeet
Your approach is certainly correct (law of sines). I get a very slightly different answer:

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

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

$\displaystyle ?=\sin^{-1}(0.6\,\sin(96^{\circ}))\approx 36.63^{\circ}.$ - Jan 24th 2011, 08:52 AMArchie Meade
- Jan 24th 2011, 08:55 AMe^(i*pi)
I got the impression the answer was prematurely rounded to 0.6 since $\displaystyle \sin(96) \approx 1$