# Use the Law of Sines to solve the triangle? 6.) A=60 degrees, a=9, c=10.

Jan 1, 2017

Check for the Ambiguous Case and, if appropriate, use the Law of Sines to solve the triangle(s).

#### Explanation:

Here is a reference for The Ambiguous Case

$\angle A$ is acute. Compute value of h:

$h = \left(c\right) \sin \left(A\right)$

$h = \left(10\right) \sin \left({60}^{\circ}\right)$

$h \approx 8.66$

$h < a < c$, therefore, two possible triangles exist, one triangle has $\angle {C}_{\text{acute}}$ and the other triangle has $\angle {C}_{\text{obtuse}}$

Use The Law of Sines to compute $\angle {C}_{\text{acute}}$

$\sin \frac{{C}_{\text{acute}}}{c} = \sin \frac{A}{a}$

$\sin \left({C}_{\text{acute}}\right) = \sin \left(A\right) \frac{c}{a}$

${C}_{\text{acute}} = {\sin}^{-} 1 \left(\sin \left(A\right) \frac{c}{a}\right)$

${C}_{\text{acute}} = {\sin}^{-} 1 \left(\sin \left({60}^{\circ}\right) \frac{10}{9}\right)$

${C}_{\text{acute}} \approx {74.2}^{\circ}$

Find the measure for angle B by subtracting the other angles from ${180}^{\circ}$:

$\angle B = {180}^{\circ} - {60}^{\circ} - {74.2}^{\circ}$

$\angle B = {45.8}^{\circ}$

Use the Law of Sines to compute the length of side b:

side $b = a \sin \frac{B}{\sin} \left(A\right)$

$b = 9 \sin \frac{{45.8}^{\circ}}{\sin} \left({60}^{\circ}\right)$

$b \approx 7.45$

For the first triangle:

$a = 9 , b \approx 7.45 , c = 10 , A = {60}^{\circ} , B \approx {45.8}^{\circ} , \mathmr{and} C \approx {74.2}^{\circ}$

Onward to the second triangle:

$\angle {C}_{\text{obtuse") ~~ 180^@ - C_("acute}}$

${C}_{\text{obtuse}} \approx {180}^{\circ} - {74.2}^{\circ} \approx {105.8}^{\circ}$

Find the measure for angle B by subtracting the other angles from ${180}^{\circ}$:

$\angle B = {180}^{\circ} - {60}^{\circ} - {105.8}^{\circ} \approx {14.2}^{\circ}$

Use the Law of Sines to compute the length of side b:

$b = 9 \sin \frac{{14.2}^{\circ}}{\sin} \left({60}^{\circ}\right)$

$b \approx 2.55$

For the second triangle:

$a = 9 , b \approx 2.55 , c = 10 , A = {60}^{\circ} , B \approx {14.2}^{\circ} , \mathmr{and} C \approx {105.8}^{\circ}$