in the triangle ABC, AC=3, BC=2, angleBAC=theta and angleABC=2theta

calculate the value of theta to the nearest tenth of a degree.

i think i'm complicating this thing for myself because i cant get farther than 2 x sin2theta = 3 x sintheta (Punch)

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- May 19th 2009, 10:10 PMthe kopitefind the angle with the greek letter
in the triangle ABC, AC=3, BC=2, angleBAC=theta and angleABC=2theta

calculate the value of theta to the nearest tenth of a degree.

i think i'm complicating this thing for myself because i cant get farther than 2 x sin2theta = 3 x sintheta (Punch) - May 19th 2009, 10:25 PMGrandad
Hello the kopiteOK, so now carry on:

$\displaystyle 2\sin2\theta = 3\sin\theta$

$\displaystyle \Rightarrow 4\sin\theta\cos\theta = 3 \sin\theta$

$\displaystyle \Rightarrow \sin\theta = 0$ or $\displaystyle \cos\theta = \tfrac34$

$\displaystyle \Rightarrow \theta = 41.4^o$

Grandad - May 19th 2009, 10:31 PMthe kopite
thanks a lot, but im here to understand. why does 2sin2theta equal 4sintheta x costheta in the first line of your reasoning?

- May 19th 2009, 10:40 PMdarthfader
I think its the 'Double Angle Identity' in Trigonometry.

Note that

$\displaystyle sin 2\theta = 2sin \theta cos\theta$

I hope this helps. (Rofl)