Length of Side b

**magentarita** · January 4th 2009, 12:36 PM

In ΔABC, m<C = 30 and a = 24. If the area of the triangle is 42, what is the length of side b?

Is this a law of sines or law of cosines problem?

**earboth** · January 5th 2009, 04:31 AM

Originally Posted by magentarita

In ΔABC, m<C = 30 and a = 24. If the area of the triangle is 42, what is the length of side b?

Is this a law of sines or law of cosines problem?

The area of a triangle is calculated by:

$A_{\Delta} = \dfrac12\cdot a\cdot b\cdot \sin(C)= \dfrac12\cdot a\cdot c\cdot \sin(B)= \dfrac12\cdot b\cdot c\cdot \sin(A)$

$\sin(30^\circ) = \dfrac12$

Plug in the values you know into the approriate equation:

$A_{\Delta} =\dfrac12\cdot 24\cdot b\cdot \sin(30^\circ)=\dfrac12\cdot 24\cdot b\cdot \dfrac12= 6b$

Therefore

$6b = 42~\implies~\boxed{b = 7}$

**magentarita** · January 5th 2009, 09:58 PM

Originally Posted by earboth

The area of a triangle is calculated by:

$A_{\Delta} = \dfrac12\cdot a\cdot b\cdot \sin(C)= \dfrac12\cdot a\cdot c\cdot \sin(B)= \dfrac12\cdot b\cdot c\cdot \sin(A)$

$\sin(30^\circ) = \dfrac12$

Plug in the values you know into the approriate equation:

$A_{\Delta} =\dfrac12\cdot 24\cdot b\cdot \sin(30^\circ)=\dfrac12\cdot 24\cdot b\cdot \dfrac12= 6b$

Therefore

$6b = 42~\implies~\boxed{b = 7}$

Yes, I forgot about those 3 formulas.

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