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ElenaW [278]
3 years ago
12

Solve the triangle. Round your answers to the nearest tenth. A. m∠A=43, m∠B=55, a=16 B. m∠A=48, m∠B=50, a=23 C. m∠A=48, m∠B=50,

a=26 D. m∠A=43, m∠B=55, a=20

Mathematics
1 answer:
alexgriva [62]3 years ago
8 0

Answer:

D. m∠A=43, m∠B=55, a=20

Step-by-step explanation:

Given:

∆ABC,

m<C = 82°

AB = c = 29

AC = b = 24

Required:

m<A, m<C, and a (BC)

SOLUTION:

Find m<B using the law of sines:

\frac{sin(B)}{b} = \frac{sin(C)}{c}

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

sin(B)*29 = sin(82)*24

\frac{sin(B)*29}{29} = \frac{sin(82)*24}{29}

sin(B) = \frac{sin(82)*24}{29}

sin(B) = 0.8195

B = sin^{-1}(0.8195)

B = 55.0

m<B = 55°

Find m<A:

m<A = 180 - (82 + 55) => sum of angles in a triangle.

= 180 - 137

m<A = 43°

Find a using the law of sines:

\frac{a}{sin(A)} = \frac{b}{sin(B)}

\frac{a}{sin(43)43} = \frac{24}{sin(55)}

Cross multiply

a*sin(55) = 25*sin(43)

a = \frac{25*sin(43)}{sin(53)}

a = 20 (approximated)

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Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

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