Question

Find all the missing parts to the triangle below.

http://curriculum.kcdistancelearning.com/courses/GEOMx-HS-A09/b/assessments/R-LawofCosinesShortAnswerQuiz/Geometry_8.4.14_Quiz_FINAL_10q.gif

Answer 1

you have to use law of cosines since you know the length of 2 sides.

it states: c^2 = a^2 + b^2 - 2ab*cosC

the angle used in cosine law is the angle opposite of the only side you don't know

c^2 = 28^2 + 15^2 -2(28)(15)cos87

c = 28.4

then you'll want to use sine law:

a/sinA = b/sinB = c/sinC

a is the side opposite to angle A, b to B, c to C.

we know c/sinC = 28.4/sin87

and it's equal to 28/sinG and also equal to 15/sinH.

You can now find angles G and H

Answer 2

So 15sinG/28=sinH?

Answer 3

yes. but you'll want to relate them to c/sinC

as you probably know: G = arcsin(28*sin87/28.4)

Answer 4

Im so lost

Answer 5

which part?

28.4/sin87 = 28/sinG = 15/sinH

you need to find G and H

Answer 6

How do I do that with the info I have?

Answer 7

u have all the info

sinG = (28*sin87)/28.4
and
sinH = (15*sin87)/28.4

Answer 8

sinG=0.98?
sinH=0.53?

Answer 9

yes. so G = arcsin(0.98456)

Answer 10

What is arcsin

Answer 11

the inverse of sin. looks like sin^-1 on your calculator. it's how you find the angle x given the value of sinx