Question

NO LINKS!!! Solve each triangle. Part 3aa

NOT A MULTIPLE CHOICE!!

1aa. B=10°, C=100°, b= 2
2bb. A= 40°, B= 40°, c= 2
3cc. A= 110°, C= 30°, c= 3​

Answer 1

Sum of interior angles of a triangle = 180°

Sine rule to find side lengths:

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

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Question 1aa

Given: B = 10°, C = 100°, b = 2

10° + 100° + A = 180°

⇒ A = 70°

`(a)/(\sin 70)=(2)/(\sin 10)=(c)/(\sin 100)`

`\implies a=\sin 70 \cdot (2)/(\sin 10)=10.82294826...`

`\implies c=\sin 100\cdot (2)/(\sin 10)=11.34256364...`

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Question 2bb

Given: A = 40°, B = 40°, c = 2

40° + 40° + C = 180°

⇒ C = 100°

`(a)/(\sin 40)=(b)/(\sin 40)=(2)/(\sin 100)`

`\implies a=\sin 40\cdot(2)/(\sin 100)=1.305407289...`

`\implies b=\sin 40\cdot(2)/(\sin 100)=1.305407289...`

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Question 1aa

Given: A = 110°, C = 30°, c = 2

110° + 30° + B = 180°

⇒ B = 40°

`(a)/(\sin 110)=(b)/(\sin 40)=(3)/(\sin 30)`

`\implies a=\sin 110 \cdot (3)/(\sin 30)=5.638155725...`

`\implies b=\sin 40\cdot(3)/(\sin 30)=3.856725658...`