# (TRIG)I need step by step help with these trig problems?

Solve the triangle with the given information

1. a=6.8

b=13.0

c.16.6

2. β=63.5°

a=12.20

c=7.80

Relevance

I assume you must find all the angles.

16.6 ^2 = 13^2 + 6.8^2 - 2(13)(6.8) cos A

275.56=169 + 46.27 - 176.8cosA

60.29 = -176.8 cos A

-.3410 = cos A

A = 110

Now do Law of Sines to get next biggest angle

16.6/sin 110 = 13/sin A

13 sin 110 = 16.6 sin A

.

.7359 = sin A

A = 47

Now to find the last angle 180-47-110 = 23

Problem 2

You can find the side opposite B by doing law of cosines

b^2 = 7.8^2 + 12.2^2 - 2(7.8)(12.2)cos63.5

b^2 = 60.84 + 148.84 - 84.92

b^2 = 124.76

b = 11.2

NOw do Law of SInes to find the second biggest angle

11.2/sine 63.5 = 12.20/sin a

sin a = .9748

a = 77

Now 180 - 63.5-77 = third angle of 39.5

• Anonymous

1. a^2 = b^2 + c^2 - 2bccosA

cos A = (b^2 + c^2 - a^2)/2bc = .923

A = 22.60 degrees.

cos B = (c^2 + a^2 - b^2)/2ca = .679

B = 47.40 degrees.

cosC = (a^2 + b^2 - c^2)/2ab = -.3515

C = 110 degrees

Check: 22.6 + 47.4 + 110 = 180

2. b^2 = c^2 + a^2 - 2ca cosB = 124.76

b = 11.17

cos A = (b^2 + c^2 - a^2)/2bc = .211

A = 77.8 deg

C = 180- 77.8 - 63.5 = 38.7 deg