# How do you solve the triangle given B=19, a=51, c=61?

Nov 2, 2016

Please see the explanation for a description of the process.

#### Explanation:

I assume that you mean that angle B = 19°

Use the Law of Cosines to find the length of side b:

$b = \sqrt{{a}^{2} + {c}^{2} - 2 \left(a\right) \left(c\right) \cos \left(B\right)}$

b = sqrt(51^2 + 61^2 - 2(51)(61)cos(19°))

b = sqrt(51^2 + 61^2 - 2(51)(61)cos(19°))

$b \approx 20.95$

Use the Law of Sines to find angle A:

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

$A = {\sin}^{-} 1 \left(\sin \left(B\right) \frac{a}{b}\right)$

A = sin^-1(sin(19°)51/20.95)

A ~~ 52°

Find angle C:

C = 180° - 19° - 52°

C = 109°

We know the lengths of all 3 sides, $a = 51 , b = 20.95 , \mathmr{and} c = 61$ and we know the measures of all 3 angles, A = 52°, B = 19° and C = 109°