Question

5. Solve the triangle in the figure.ООA. BC = 9; m ZA = 36.9°; mZB = 53.1°; m2C = 90°B. BC = 11; mZA = 42.8°; m ZB = 47.2°; mZC = 90°C. BC = 9; m_A = 53.1°; mZB = 6.9°; m2C = 90°D. BC = 19; mZA = 36.9°; m ZB = 53.1°; mZC = 90°

Answer 1

We are given the right angle triangle.

First, let us find the length of side BC using the Pythagorean theorem.

`\begin{gathered} AC^2+BC^2=AB^2 \\ 12^2+BC^2=15^2 \\ BC^2=15^2-12^2 \\ BC^2=225-114 \\ BC^2=81 \\ BC=\sqrt[]{81} \\ BC=9 \end{gathered}`

Therefore, the length of side BC = 9

The angle m∠C is clearly 90°

Let us find the angle m∠A and m∠B

We can use the law of sines to find the m∠A and m∠B

`(B)/(\sin(b))=(C)/(\sin(c))`

Let us substitute the values and find m∠B

`\begin{gathered} (12)/(\sin (b))=(15)/(\sin (90\degree)) \\ (12)/(\sin (b))=(15)/(1) \\ \sin (b)=(12)/(15) \\ b=\sin ^(-1)((12)/(15))_{} \\ b=53.1\degree \end{gathered}`

So the angle m∠B is 53.1°

Now we have two angles so the third angle can be found by

`\begin{gathered} m\angle A+m\angle B+m\angle C=180\degree \\ m\angle A+53.1\degree+90\degree=180\degree \\ m\angle A=180\degree-53.1\degree-90\degree \\ m\angle A=36.9\degree \end{gathered}`

Therefore, the angles are

BC = 9

m∠A = 36.9°

m∠B = 53.1°

m∠C = 90°

Therefore, the correct option is A