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

`Side\ B = 6.0`

`\alpha = 56.3`

`\theta = 93.7`

Explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with
`\alpha, \beta, \theta`

[See Attachment for Triangle]

`A = 10cm`

`C = 12cm`

`\beta = 30`

What the question is to calculate the third length (Side B) and the other 2 angles (
`\alpha\ and\ \theta`)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

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

Substitute:
`A = 10`,
`C =12`;
`\beta = 30`

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

`B^2 = 100 + 144 - 240*0.86602540378`

`B^2 = 100 + 144 - 207.846096907`

`B^2 = 36.153903093`

Take Square root of both sides

`√(B^2) = √(36.153903093)`

`B = √(36.153903093)`

`B = 6.0128115797`

`B = 6.0` (Approximated)

Calculating Angle
`\alpha`

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

Substitute:
`A = 10`,
`C =12`;
`B = 6`

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

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

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

`100 = 180 - 144 *Cos\alpha`

Subtract 180 from both sides

`100 - 180 = 180 - 180 - 144 *Cos\alpha`

`-80 = - 144 *Cos\alpha`

Divide both sides by -144

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

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

`0.5555556 = Cos\alpha`

Take arccos of both sides

`Cos^(-1)(0.5555556) = Cos^(-1)(Cos\alpha)`

`Cos^(-1)(0.5555556) = \alpha`

`56.25098078 = \alpha`

`\alpha = 56.3` (Approximated)

Calculating
`\theta`

Sum of angles in a triangle = 180

Hence;

`\alpha + \beta + \theta = 180`

`30 + 56.3 + \theta = 180`

`86.3 + \theta = 180`

Make
`\theta` the subject of formula

`\theta = 180 - 86.3`

`\theta = 93.7`