Question

Solve the triangle. Round your answers to the nearest tenth. Law of sines 6 A. M∠C=34 m ∠ C = 34 , b=25 b = 25 , c=16 c = 16 B. M∠C=34 m ∠ C = 34 , b=25 b = 25 , c=17 c = 17 C. M∠C=34 m ∠ C = 34 , b=22 b = 22 , c=17 c = 17 D. M∠C=34 m ∠ C = 34 , b=27.9 b = 27.9 , c=16 c = 16

Answer 1

Answer:

$A = 43.0^o$

$B = 55.0^o$

$BC = 20.0$

Step-by-step explanation:

I will solve this question with the attached triangle

From the attachment, we have:

$\angle C =82^o$

$AB = 29$

$AC = 24$

Required

Solve the triangle

First, we calculate $\angle B$ using sine law:

$\frac{AC}{\sin(B)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{24}{\sin(B)} = \frac{29}{\sin(82^o)}$

$\frac{24}{\sin(B)} = \frac{29}{0.9903}$

Cross multiply

$\sin(B) * 29 = 24 * 0.9903$

$\sin(B) * 29 = 23.7672$

Divide both sides by 29

$\sin(B) = 0.8196$

Take arcsin of both sides

$B = \sin^{-1}(0.8196)$

$B = 55.0^o$

Next, calculate $\angle A$ using:

$A + B + C = 180^o$ --- angles in a triangle

$A + 55^o + 82^o = 180^o$

Collect like terms

$A = 180^o-55.0^o - 82^o$

$A = 43.0^o$

Next, calculate BC using sine laws

$\frac{BC}{\sin(A)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{BC}{\sin(43.0)} = \frac{29}{\sin(82)}$

$\frac{BC}{0.6820} = \frac{29}{0.9903}$

Make BC the subject

$BC = \frac{29*0.6820}{0.9903}$

$BC = 20.0$