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

For questions 1 - 6, solve the given triangles by finding the missing angle and other-example-1
User Vaviloff
by
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2 Answers

3 votes

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

User Quentino
by
7.2k points
2 votes

Answer:

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)

User Timss
by
6.1k points
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