Question

Given: △ABC, AB=5sqrt2 m∠A=45°, m∠C=30° Find: BC and AC

Answer 1

BC is 10 units and AC is $5+5\sqrt{3}$ units

Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

• $\frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}$
• AB is opposite to ∠C
• BC is opposite to ∠A
• AC is opposite to ∠B

Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

∵ m∠A + m∠B + m∠C = 180

∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ $\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

∵ AB = $5\sqrt{2}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}$

- By using cross multiplication

∴ (BC) × sin(30) = $5\sqrt{2}$ × sin(45)

∵ sin(30) = 0.5 and sin(45) = $\frac{1}{\sqrt{2}}$

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ $\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

- Substitute AB and the 3 angles in the rule above

∴ $\frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}$

- By using cross multiplication

∴ (AC) × sin(30) = $5\sqrt{2}$ × sin(105)

∵ sin(105) = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∴ 0.5 (AC) = $\frac{5+5\sqrt{3}}{2}$

- Divide both sides by 0.5

∴ AC = $5+5\sqrt{3}$ units

BC is 10 units and AC is $5+5\sqrt{3}$ units

Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

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