Question

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. (5 points)

A = 56°, a = 12, b = 14

B = 75.3°, C = 48.7°, c = 10.9; B = 104.7°, C = 19.3°, c = 4.8
B = 14.7°, C = 109.3°, c = 13.7; B = 165.3°, C = 70.7°, c = 13.7
B = 75.3°, C = 48.7°, c = 13.2; B = 104.7°, C = 19.3°, c = 13.2
B = 14.7°, C = 109.3°, c = 10.5; B = 165.3°, C = 70.7°, c = 10.5

Answer 1

@SolomonZelman @ganeshie8 @AlexandervonHumboldt2

Answer 2

@mathstudent55

Answer 3

Do you know the law of sines?

Answer 4

I have heard of it only

Answer 5

it is a/sinA=b/sinB=c/sinC, right?

Answer 6

Is that right @mathstudent55

Answer 7

Yes. That is it.

Answer 8

The law if sines states that the ratio of the length of a side of a triangle to the sine of the opposite angle is equal for all sides and opposite angles.

Answer 9

so to find the angle B I need to solve sinB=b*sinA/a

Answer 10

In order to use the law of sines, you need to find that ratio.
In order to do that, you must know the length of one side and the measure of the opposite angle to that side.

Answer 11

sinB=14*sin56^deg/12 it is about 75.3 degrees.

Answer 12

Look at the given info.
You are told A = 56 deg, and a = 12
That is what you need to find the ratio.
Then since you are given one other side length, you can find the opposite angle.

Answer 13

c=12*sin48.7deg/sin56deg

Answer 14

ok, is the correct answer A?

Answer 15

\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \)

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

\(\dfrac{12}{\sin 56^o} = \dfrac{14}{\sin B} \)

\(\sin B = \dfrac{14\sin 56^o}{12} \)

\(B = \sin^{-1} \dfrac{14\sin 56^o}{12} \)

\(B \approx 75.3^o\)

You are correct.

Answer 16

Now you find C:

C = 180 - 75.3 - 56 = 48.7 deg

and you do the law of sines for c and C

Answer 17

\(\dfrac{a}{\sin A} = \dfrac{c}{\sin C}\)

\(\dfrac{12}{\sin 56^o} = \dfrac{c}{\sin 48.7^o}\)

\(c = \dfrac{12\sin 48.7^o}{\sin 56^o} \)

\(c = 10.9\)

Answer 18

Yes, the answer is A.