hi, i need to find the missing angle from this shape:

Attachment 20478

im not really sure how to do this so if someone could tell me i will be very grateful!

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- Jan 17th 2011, 08:17 AMandyboy179find missing angle?
hi, i need to find the missing angle from this shape:

Attachment 20478

im not really sure how to do this so if someone could tell me i will be very grateful! - Jan 17th 2011, 08:27 AMe^(i*pi)
Have you covered the cosine rule?

$\displaystyle c^2=a^2+b^2-2ab\cos C$ which may also be written as $\displaystyle \cos C = \dfrac{a^2+b^2-c^2}{2ab}$

Edit: Misread angle as a side - use the law of sines instead (post left here for continuity) - Jan 17th 2011, 08:39 AMandyboy179
would c^2= 6, a^2= 10 and b^2= 96??

- Jan 17th 2011, 08:50 AMskeeter
Law of sines ...

$\displaystyle \displaystyle \frac{\sin(96^\circ)}{10} = \frac{\sin{\theta}}{6}

$

solve for $\displaystyle \theta$

http://www.mathhelpforum.com/math-he...ngle-shape.jpg - Jan 17th 2011, 08:51 AMe^(i*pi)
My apologies, I misread that angle as a side. Instead you can use the sine rule

$\displaystyle \dfrac{\sin A}{a} = \dfrac{\sin B}{b}$

In this case you have

- $\displaystyle a = 6$
- $\displaystyle \sin B = 96$
- $\displaystyle b = 10$

$\displaystyle \sin A = \dfrac{a \sin B}{b} = \dfrac{6 \sin(96)}{10}$ - Jan 17th 2011, 10:38 AMandyboy179
would i start it like this?

Attachment 20481 - Jan 17th 2011, 10:43 AMe^(i*pi)
Yes, that's fine.

Me and skeeter used the other way of writing it which is the same as you but flipped over. I have started it off in post 6 - Jan 17th 2011, 10:48 AMandyboy179
would i then do sin?= 6x10/sin96=60.33?