Question

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

Answer 1

Answer:

c = 13.2

Step-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:

# $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

* 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°

∵ $\frac{4}{sin(16)}=\frac{c}{sin(115)}$

- By using cross multiplication

∴ c sin(16) = 4 sin(115) ⇒ divide both sides by sin(16)

∴ $c=\frac{4(sin115)}{sin16}=13.2$

* c = 13.2