Question

For questions 1 - 6, solve the given triangles by finding the missing angle and other side lengths please help!!!

Answer 1

1. The missing angle in the given triangle is 28°

2. The missing lengths in the given triangle are:

• Side a = 28
• Side c = 19

How to calculate the missing angle and lengths?

1. The missing angle can be calculated as follow:

40 + 112 + Missing angle = 180 (sum of angle in a triangle)

152 + Missing angle = 180

Missing angle = 180 - 152

Missing angle = 28°

2. The missing lengths can be calculated as shown below:

For side a:

• Angle A = 112°
• Angle B = 28°
• Side b = 14
• Side a =?

`(a)/(SineA)\ = (b)/(SineB)\\\\(a)/(Sine\ 112)\ = (14)/(Sine\ 28)\\\\a\ * Sine\ 28 = 14\ * Sine\ 112\\\\a = (14\ * Sine\ 112)/(Sine\ 28) \\\\a \approx 28`

For side c:

• Angle C = 40°
• Angle B = 28°
• Side b = 14
• Side c =?

`(c)/(SineC)\ = (b)/(SineB)\\\\(c)/(Sine\ 40)\ = (14)/(Sine\ 28)\\\\c\ * Sine\ 28 = 14\ * Sine\ 40\\\\c = (14\ * Sine\ 40)/(Sine\ 28) \\\\c = 19`

Answer 2

m<B = 28°

`AB = 19.2`

`AC = 27.6`

Explanation:

Given:

m<A = 112°

m<C = 40°

AC = 14

Required:

Find, m<B, AB and BC.

SOLUTION:

✍️m<B = 180° - (112° + 40°) (sum of ∆)

m<B = 28°

✍️Using sine rule, find AB.

`(AB)/(sin(C)) = (AC)/(sin(B))`

Plug in the values into the equation.

`(AB)/(sin(40)) = (14)/(sin(28))`

Cross multiply

`AB*sin(28) = 14*sin(40)`

Divide both sides by sin(28)

`AB = (14*sin(40))/(sin(28))`

`AB = 19.2` (nearest tenth)

✍️Using sine rule, find BC.

`(BC)/(sin(A)) = (AC)/(sin(B))`

Plug in the values into the equation.

`(BC)/(sin(112)) = (14)/(sin(28))`

Cross multiply

`AC*sin(28) = 14*sin(112)`

Divide both sides by sin(28)

`AC = (14*sin(112))/(sin(28))`

`AC = 27.6` (nearest tenth)