# How do you solve the right triangle ABC given A = 61°, a = 17, b = 19?

##### 2 Answers
Jun 8, 2015

If ABC is a right triangle
$\textcolor{w h i t e}{\text{XXXX}}$with $A = {61}^{\circ}$
$\textcolor{w h i t e}{\text{XXXX}}$and $a = 17$
$\textcolor{w h i t e}{\text{XXXX}}$and $b = 19$
It might (at first appear that 2 configurations are possible:

$\angle B = {90}^{\circ}$ (since it is a right triangle)
$\angle C = {29}^{\circ}$ (since $\angle A + \angle B + \angle C = {180}^{\circ}$
and
$c = \sqrt{{19}^{2} - {17}^{2}}$ (approximately $8.485$)

Unfortunately
Attempting to use the Law of Sines to verify these results:
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

we get
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\sin A}{a} = \sin \frac{{61}^{\circ}}{17} = 0.051448$
and
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\sin B}{b} = \sin \frac{{90}^{\circ}}{19} = 0.052362$
and
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{\sin C}{c} = \sin \frac{{29}^{\circ}}{8.485} = 0.057135$

Conclusion
The given values are not those of a right triangle!

Jun 8, 2015

No solution is possible with the given data.