Question

NO LINKS!!! Solve each triangle. Part 2a

NOT A MULTIPLE CHOICE

1a. A= 40°, B= 20, a= 2
2b. A= 50°, C= 20°, a= 3
3c. B= 70°, C= 10°, b= 5
4d. A= 70°, B= 60°, c= 4

​

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)`

----------------------------------------------------------------------

Question 1a

Given: A = 40°, B = 20°, a = 2

40° + 20° + C = 180°

⇒ C = 120°

`(2)/(\sin 40)=(b)/(\sin 20)=(c)/(\sin 120)`

`\implies b=\sin20 \cdot(2)/(\sin 40)=1.064177772...`

`\implies c=\sin 120 \cdot (2)/(\sin 40)=2.694592711...`

----------------------------------------------------------------------

Question 2b

Given: A = 50°, C = 20°, a = 3

50° + 20° + B = 180°

⇒ B = 110°

`(3)/(\sin 50)=(b)/(\sin 110)=(c)/(\sin 20)`

`\implies b=\sin 110 \cdot(3)/(\sin 50)=3.680044791...`

`\implies c=\sin 20 \cdot(3)/(\sin 50)=1.339426765...`

----------------------------------------------------------------------

Question 3c

Given: B = 70°, C = 10°, b = 5

70° + 10° + A = 180°

⇒ A = 100°

`(a)/(\sin 100)=(5)/(\sin 70)=(c)/(\sin 10)`

`\implies a=\sin 100 \cdot (5)/(\sin 70)=5.240052605...`

`\implies c=\sin 10 \cdot (5)/(\sin 70)=0.9239626545...`

----------------------------------------------------------------------

Question 4d

Given: A = 70°, B = 60°, c = 4

70° + 60° + C = 180°

⇒ C = 50°

`(a)/(\sin 70)=(b)/(\sin 60)=(4)/(\sin 50)`

`\implies a=\sin 70 \cdot (4)/(\sin 50)=4.906726388...`

`\implies b=\sin 60 \cdot (4)/(\sin 50)=4.522063499...`