In △ABC, m∠A=16°, m∠B=49°, and a=4. Find c to the nearest tenth.NOT A RIGHT TRIANGLE

Answer:c = 13.2Step-by-step explanation:* Lets explain how to solve the problem- In Δ ABC # Side a is opposite to ∠A# Side b is opposite to ∠B# Side c is opposite to ∠C- The sine rule is:# [tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]* Lets solve the problem- In Δ ABC∵ m∠A = 16°∵ m∠B = 49°∵ The sum of the measures of the interior angles of a triangle is 180°∴ m∠A + m∠B + m∠C = 180°∴ 16° + 49° + m∠C = 180°∴ 65° + m∠C = 180° ⇒ subtract 65° from both sides∴ m∠C = 115°- Lets use the sine rule to find c∵ a = 4 and m∠A = 16°∵ m∠C = 115°∵ [tex]\frac{4}{sin(16)}=\frac{c}{sin(115)}[/tex]- By using cross multiplication∴ c sin(16) = 4 sin(115) ⇒ divide both sides by sin(16)∴ [tex]c=\frac{4(sin115)}{sin16}=13.2[/tex]* c = 13.2