Question

Solve the triangle. Round your answers to the nearest tenth.

Answer 1

Angles:
`\angle A = 43^(\circ)`
`\angle B = 55^(\circ)`
`\angle C = 82^(\circ)`

Sides:
`BC = 20`
`AB = 29`
`AC =24`

Explanation:

See attachment for complete question

From the attachment, we have:

`AB = 29`

`AC =24`

`\angle C = 82^(\circ)`

Required

Complete the missing side and missing angles

To calculate angle B, we apply sine laws:

`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)`

In this case:

`(AB)/(sinC)=(AC)/(sinB)`

This gives:

`(29)/(sin(82^(\circ)))=(24)/(sinB)`

`(29)/(0.9903)=(24)/(sinB)`

Cross Multiply

`sinB * 29 = 24 * 0.9903`

Divide both sides by 29

`sinB = (24 * 0.9903)/(29)`

`sinB = 0.8196`

Take arcsin of both sides

`B = sin^(-1)(0.8196)`

`B = 55^(\circ)`

So:

`\angle B = 55^(\circ)`

To solve for the third angle, we make use of:

`\angle A + \angle B + \angle C = 180^(\circ)`

This gives:

`\angle A + 55^(\circ) + 82^(\circ) = 180^(\circ)`

`\angle A + 137^(\circ)= 180^(\circ)`

`\angle A = 180^(\circ)- 137^(\circ)`

`\angle A = 43^(\circ)`

Hence, the angles are:

`\angle A = 43^(\circ)`
`\angle B = 55^(\circ)`
`\angle C = 82^(\circ)`

To calculate the length of the third side, we apply cosine law

`BC^2 = AB^2 + AC^2 - 2*AB*AC*cosA`

`BC^2 = 29^2 + 24^2 - 2*29*24*cos(43^(\circ))`

`BC^2 = 841+ 576 - 1392*cos(43^(\circ))`

`BC^2 = 841+ 576 - 1392*0.7314`

`BC^2 = 841+ 576 - 1018.11`

`BC^2 = 398.89`

Take the square root of both sides

`BC = \sqrt{398.89`

`BC = 19.9722307217`

`BC = 20`