In △ABC, m∠A=57°, m∠B=37°, and a=11. Find c to the nearest tenth.

Answer:c = 13.1Step-by-step explanation:* Lets explain how to solve the problem- In Δ ABC # ∠A is opposite to side a# ∠B is opposite to side b# ∠C is opposite to side c- The sine rule is:# [tex]\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}[/tex]* Lets solve the problem- In Δ ABC∵ m∠A = 57°∵ m∠B = 37°∵ The sum of the measures of the interior angles of a triangle is 180°∴ m∠A + m∠B + m∠C = 180°∴ 57° + 37° + m∠C = 180°∴ 94° + m∠C = 180° ⇒ subtract 94° from both sides∴ m∠C = 86°- Lets use the sine rule to find c∵ a = 11 and m∠A = 57°∵ m∠C = 86°∵ [tex]\frac{sin(57)}{11}=\frac{sin(86)}{c}[/tex]- By using cross multiplication∴ c sin(57) = 11 sin(86) ⇒ divide both sides by sin(57)∴ [tex]c=\frac{11(sin86)}{sin57}=13.1[/tex]* c = 13.1