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

Answer:

The missing side is $B = 6.0\ cm$

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

Step-by-step 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

$B = \sqrt{36.2}$

$B = 6.0$ (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

$\frac{-80}{-144} = Cos\alpha$

$0.5556 = Cos\alpha$

$\alpha = cos^{-1}(0.5556)$

$\alpha = 56.2$ (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

$\frac{8}{-120} = Cos\theta$

$-0.0667= Cos\theta$

$\theta = cos^{-1}(-0.0667)$

$\theta = 93.8$ (Approximated)