Question

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

Answer 1

Answer:

c = 13.1

Step-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:

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

* 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°

∵ $\frac{sin(57)}{11}=\frac{sin(86)}{c}$

- By using cross multiplication

∴ c sin(57) = 11 sin(86) ⇒ divide both sides by sin(57)

∴ $c=\frac{11(sin86)}{sin57}=13.1$

* c = 13.1