# find missing angle?

• Jan 17th 2011, 08:17 AM
andyboy179
find 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 AM
e^(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 AM
andyboy179
would c^2= 6, a^2= 10 and b^2= 96??
• Jan 17th 2011, 08:50 AM
skeeter
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 AM
e^(i*pi)
Quote:

Originally Posted by andyboy179
would c^2= 6, a^2= 10 and b^2= 96??

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 AM
andyboy179
would i start it like this?

Attachment 20481
• Jan 17th 2011, 10:43 AM
e^(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 AM
andyboy179
would i then do sin?= 6x10/sin96=60.33?