Question

consider a triangle (abc) like the one below. suppose that (a = 52), (b = 64), and (a=30^{circ}). (the figure is not drawn to scale.) solve the triangle. carry your intermediate computations to at least four - decimal places, and round your answers to the nearest tenth. if no such triangle exists, enter
o solution.\ if there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, $\frac{a}{\sin A}=\frac{b}{\sin B}$. Substitute $a = 52$, $b = 64$, and $A=30^{\circ}$ into the formula: $\sin B=\frac{b\sin A}{a}=\frac{64\times\sin30^{\circ}}{52}=\frac{64\times0.5}{52}=\frac{32}{52}\approx0.6154$.

Step2: Find angle B

Since $\sin B\approx0.6154$, then $B=\sin^{- 1}(0.6154)\approx38.0^{\circ}$ or $B = 180^{\circ}-38.0^{\circ}=142.0^{\circ}$. But when $B = 142.0^{\circ}$, $A + B=30^{\circ}+142.0^{\circ}=172.0^{\circ}<180^{\circ}$.

Step3: Find angle C for the first - case of B

When $B\approx38.0^{\circ}$, $C=180^{\circ}-A - B=180^{\circ}-30^{\circ}-38.0^{\circ}=112.0^{\circ}$.

Step4: Use the Law of Sines to find c for the first - case of B

Again, by the Law of Sines $\frac{a}{\sin A}=\frac{c}{\sin C}$. So $c=\frac{a\sin C}{\sin A}=\frac{52\times\sin112.0^{\circ}}{\sin30^{\circ}}=\frac{52\times0.9272}{0.5}\approx96.4$.

Step5: Find angle C for the second - case of B

When $B = 142.0^{\circ}$, $C=180^{\circ}-A - B=180^{\circ}-30^{\circ}-142.0^{\circ}=8.0^{\circ}$.

Step6: Use the Law of Sines to find c for the second - case of B

By the Law of Sines $\frac{a}{\sin A}=\frac{c}{\sin C}$, so $c=\frac{a\sin C}{\sin A}=\frac{52\times\sin8.0^{\circ}}{\sin30^{\circ}}=\frac{52\times0.1392}{0.5}\approx14.4$.

Answer:

$B\approx38.0^{\circ},C\approx112.0^{\circ},c\approx96.4$ or $B\approx142.0^{\circ},C\approx8.0^{\circ},c\approx14.4$