Question

Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Round all answers using one decimal place.

Answer 1

The missing side is
`B = 6.0\ cm`

The missing angles are
`\alpha = 56.2` and
`\theta = 93.8`

Explanation:

Given

`A = 10\ cm`

`C = 12\ cm`

`\beta = 30`

The implication of this question is to solve for the missing side and the two missing angles

Represent

Angle A with
`\alpha`

Angle B with
`\beta`

Angle C with
`\theta`

Calculating B

This will be calculated using cosine formula as thus;

`B^2 = A^2 + C^2 - 2ACCos\beta`

Substitute values for A, C and
`\beta`

`B^2 = 10^2 + 12^2 - 2 * 10 * 12 * Cos30`

`B^2 = 100 + 144 - 240 * 0.8660`

`B^2 = 100 + 144 - 207.8`

`B^2 = 36.2`

Take Square root of both sides

`B = √(36.2)`

`B = 6.0` (Approximated)

Calculating
`\alpha`

This will be calculated using cosine formula as thus;

`A^2 = B^2 + C^2 - 2BCCos\alpha`

Substitute values for A, B and C

`A^2 = B^2 + C^2 - 2BCCos\alpha`

`10^2 = 6^2 + 12^2 - 2 * 6 * 12 * Cos\alpha`

`100 = 36 + 144 - 144Cos\alpha`

Collect Like Terms

`100 - 36 - 144 = -144Cos\alpha`

`-80 = -144Cos\alpha`

Divide both sides by -144

`(-80)/(-144) = Cos\alpha`

`0.5556 = Cos\alpha`

`\alpha = cos^(-1)(0.5556)`

`\alpha = 56.2` (Approximated)

Calculating
`\theta`

This will be calculated using cosine formula as thus;

`C^2 = B^2 + A^2 - 2BACos\theta`

Substitute values for A, B and C

`12^2 = 6^2 + 10^2 - 2 * 6 * 10Cos\theta`

`144 = 36 + 100 - 120Cos\theta`

Collect Like Terms

`144 - 36 - 100 = -120Cos\theta`

`8 = -120Cos\theta`

Divide both sides by -120

`(8)/(-120) = Cos\theta`

`-0.0667= Cos\theta`

`\theta = cos^(-1)(-0.0667)`

`\theta = 93.8` (Approximated)