# Thread: Length of Side b

1. ## Length of Side b

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?

2. 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:

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

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

Plug in the values you know into the approriate equation:

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

Therefore

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

3. ## yes........

Originally Posted by earboth
The area of a triangle is calculated by:

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

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

Plug in the values you know into the approriate equation:

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

Therefore

$\displaystyle 6b = 42~\implies~\boxed{b = 7}$
Yes, I forgot about those 3 formulas.