BC is 10 units and AC is [tex]5+5\sqrt{3}[/tex] units
Step-by-step explanation:
Let us revise the sine rule
In ΔABC:
- [tex]\frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}[/tex]
- 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°
∵ [tex]\frac{AB}{sin(C)}=\frac{BC}{sin(A)}[/tex]
∵ AB = [tex]5\sqrt{2}[/tex]
- Substitute AB and the 3 angles in the rule above
∴ [tex]\frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}[/tex]
- By using cross multiplication
∴ (BC) × sin(30) = [tex]5\sqrt{2}[/tex] × sin(45)
∵ sin(30) = 0.5 and sin(45) = [tex]\frac{1}{\sqrt{2}}[/tex]
∴ 0.5 (BC) = 5
- Divide both sides by 0.5
∴ BC = 10 units
∵ [tex]\frac{AB}{sin(C)}=\frac{AC}{sin(B)}[/tex]
- Substitute AB and the 3 angles in the rule above
∴ [tex]\frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}[/tex]
- By using cross multiplication
∴ (AC) × sin(30) = [tex]5\sqrt{2}[/tex] × sin(105)
∵ sin(105) = [tex]\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
∴ 0.5 (AC) = [tex]\frac{5+5\sqrt{3}}{2}[/tex]
- Divide both sides by 0.5
∴ AC = [tex]5+5\sqrt{3}[/tex] units
BC is 10 units and AC is [tex]5+5\sqrt{3}[/tex] units
Learn more:
You can learn more about the sine rule in brainly.com/question/12985572
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