Solve triangle problems: 1. In triangle ABC, ABC=80, CAB=45, AC=50. Find BC. 2. In triangle PQR, RPQ=60…

Question

Solve triangle problems: 1. In triangle ABC, ABC=80, CAB=45, AC=50. Find BC. 2. In triangle PQR, RPQ=60, QRP=50, RP=6.5. Find PQ.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions for problems 1 and 2 again, based on the values in the image:

1. In the triangle ABC, ABC = 80, CAB = 45 and AC = 50. Find BC.

Find the third angle of the triangle.
The sum of angles in a triangle is $180^\circ$ .
$\angle ACB = 180^\circ - 80^\circ - 45^\circ = 55^\circ$
Apply the Sine Rule.
$\frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)}$ $\frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)}$
Solve for BC.
$BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)}$ $BC = \frac{50 \times 0.7071}{0.9848}$ $BC \approx 35.900$
The length of BC is approximately $\boxed{\text{35.90 units}}$ .

2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 6.5. Find PQ.
Find the third angle of the triangle.
$\angle PQR = 180^\circ - 60^\circ - 50^\circ = 70^\circ$
Apply the Sine Rule.
$\frac{PQ}{\sin(\angle QRP)} = \frac{RP}{\sin(\angle PQR)}$ $\frac{PQ}{\sin(50^\circ)} = \frac{6.5}{\sin(70^\circ)}$
Solve for PQ.
$PQ = \frac{6.5 \times \sin(50^\circ)}{\sin(70^\circ)}$ $PQ = \frac{6.5 \times 0.7660}{0.9397}$ $PQ \approx 5.298$
The length of PQ is approximately $\boxed{\text{5.30 units}}$ .

Solve triangle problems: 1. In triangle ABC, ABC=80, CAB=45, AC=50. Find BC. 2. In triangle PQR, RPQ=60, QRP=50, RP=6.5. Find PQ.