Question

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

Answer 1

Answer:

Angles: $\angle A = 43^{\circ}$   $\angle B = 55^{\circ}$    $\angle C = 82^{\circ}$

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

Step-by-step 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:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In this case:

$\frac{AB}{sinC}=\frac{AC}{sinB}$

This gives:

$\frac{29}{sin(82^{\circ})}=\frac{24}{sinB}$

$\frac{29}{0.9903}=\frac{24}{sinB}$

Cross Multiply

$sinB * 29 = 24 * 0.9903$

Divide both sides by 29

$sinB = \frac{24 * 0.9903}{29}$

$sinB = 0.8196$

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

$BC = \sqrt{398.89$

$BC = 19.9722307217$

$BC = 20$