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Svetlanka [38]
3 years ago
10

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
Mathematics
1 answer:
PolarNik [594]3 years ago
6 0

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

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