How do you find the angles of a triangle given sides a=8, b=10, c=12?

Aug 27, 2016

$\hat{A} = {41.41}^{\circ} , \hat{B} = {55.77}^{\circ} , \hat{C} = {82.82}^{\circ}$

Explanation:

With the help of sinus law that establishes

$\frac{a}{\sin} \left(\hat{A}\right) = \frac{b}{\sin} \left(\hat{B}\right) = \frac{c}{\sin} \left(\hat{C}\right)$

and also

$\hat{A} + \hat{B} + \hat{C} = \pi$

The formulation is

{ (a/sin(hat A) = b/sin(hat B)), (b/sin(hat B) = c/sin(hat C)), (hatA + hatB + hatC = pi) :}

By knowing that

$\sin \left(\pi - \left(\hat{A} + \hat{B}\right)\right) = \sin \left(\hat{A} + \hat{B}\right)$

This system can be reduced to

{ (a/sin(hat A) = b/sin(hat B)), (b/sin(hat B) = c/sin(hat A + hat B)) :}

Solving for $\hat{A} , \hat{B}$ after substituting

$\sin \left(\hat{A} + \hat{B}\right) = \sin \left(\hat{A}\right) \cos \left(\hat{B}\right) + \sin \left(\hat{B}\right) \cos \left(\hat{A}\right)$
and $\cos \left(\alpha\right) = \sqrt{1 - \sin {\left(\alpha\right)}^{2}}$ we have:

$\hat{A} = \arctan \left(\frac{{b}^{2} + {c}^{2} - {a}^{2}}{b c} , \frac{\sqrt{\left(b + c - a\right) \left(a + b - c\right) \left(a - b + c\right) \left(a + b + c\right)}}{b c}\right)$
hat B = arctan((a^2-b^2+c^2)/(ac),sqrt((b+c-a)(a+b-c)(a-b+c)(a+b+c))/(ac)

or substituting values

$\hat{A} = \arctan \left(\frac{\sqrt{7}}{3}\right) = {41.41}^{\circ} , \hat{B} = \arctan \left(5 \frac{\sqrt{7}}{9}\right) = {55.77}^{\circ}$ and $\hat{C} = {180}^{\circ} - \left({41.41}^{\circ} + {55.77}^{\circ}\right) = {82.82}^{\circ}$