# How do you solve the triangle ABC given A=30, b=4, c=6?

Feb 20, 2017

$a = 2 \sqrt{13 - 6 \sqrt{3}} \cong 3.23$

hat B=sin^(-1)(1/(sqrt(13-6sqrt3)))~=38.26°

hat C~=180°-hatA-hatB~=111.74°

#### Explanation:

You would apply the cosine theorem to find the third side a:

a=sqrt(b^2+c^2-2bc cos hatA

Then

a=sqrt(4^2+6^2-2*4*6*cos30°)

$= \sqrt{16 + 36 - {\cancel{48}}^{24} \cdot \frac{\sqrt{3}}{\cancel{2}}}$

=sqrt(4(4+9-6sqrt(3))

$= 2 \sqrt{13 - 6 \sqrt{3}} \cong 3.23$

To find the angle $\hat{B}$, you would apply the sine theorem:

$\frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}}$

or

$\sin \hat{B} = \frac{b}{a} \cdot \sin \hat{A}$

=cancel4^2/(cancel2sqrt(13-6sqrt3))*sin 30°

$\frac{\cancel{2}}{\sqrt{13 - 6 \sqrt{3}}} \cdot \frac{1}{\cancel{2}}$

$= \frac{1}{\sqrt{13 - 6 \sqrt{3}}}$

then

hat B=sin^(-1)(1/(sqrt(13-6sqrt3)))~=38.26°

hat C~=180°-hatA-hatB~=111.74°