# How do you find all possible measures of B if A = 30^\circ, a = 13, b = 15 for triangle ABC?

Apr 14, 2018

See below.

#### Explanation:

From the diagram you can see that we have an ambiguous case. This gives us to unique triangles:

ABC and ACD

Using the sine rule:

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

Solving for $\angle A B C$

$\sin \frac{30}{13} = \sin \frac{B}{15}$

Rearranging:

$\sin \left(B\right) = \frac{15 \sin \left(30\right)}{13} = \frac{\left(15\right) \left(\frac{1}{2}\right)}{13} = \frac{15}{26}$

$B = \arcsin \left(\frac{15}{26}\right) = {35.23}^{\circ}$ 2 d.p.

But we also have another angle $\angle A D C$

These are angles on a straight line so:

$\angle A D C = {180}^{\circ} - \arcsin \left(\frac{15}{26}\right) = {144.77}^{\circ}$ 2 d.p.

Triangle ABC$\left[1\right]$

Using $B = {35.23}^{\circ}$

$\angle A C B = {180}^{\circ} - \left({30}^{\circ} + {35.23}^{\circ}\right) = {114.77}^{\circ}$

Side $c$:

$\sin \frac{{30}^{\circ}}{13} = \sin \frac{114.77}{c}$

$c = \frac{13 \sin \left(114.77\right)}{\sin \left({30}^{\circ}\right)} = 23.61$

So triangle ABC$\left[1\right]$

$\boldsymbol{A} = {30}^{\circ}$

$\boldsymbol{B} = {35.23}^{\circ}$

$\boldsymbol{C} = {114.77}^{\circ}$

$\boldsymbol{a} = 13$

$\boldsymbol{b} = 15$

$\boldsymbol{c} = 23.61$

Triangle ABC$\left[2\right]$

Using $B = {144.77}^{\circ}$

$\angle A C B$

${180}^{\circ} - \left({114.77}^{\circ} + {30}^{\circ}\right) = {5.23}^{\circ}$

side $c$

$\sin \frac{{30}^{\circ}}{13} = \sin \frac{{5.23}^{\circ}}{c}$

$c = \frac{13 \sin \left({5.23}^{\circ}\right)}{\sin \left({30}^{\circ}\right)} = 2.37$

So triangle ABC$\left[2\right]$

$\boldsymbol{A} = {30}^{\circ}$

$\boldsymbol{B} = {144.77}^{\circ}$

$\boldsymbol{C} = {5.23}^{\circ}$

$\boldsymbol{a} = 13$

$\boldsymbol{b} = 15$

$\boldsymbol{c} = 2.37$