Need help finding the missing angles and sides for a triangle?

Question

I'm having trouble finding the missing sides and angles for the triangle. I know you have to use law of sines to find them, but I'm having trouble. Any help is greatly appreciated!

 

A= 42°, C= 96°, c= 12 in

 

 

Thanks in advance!

 

 

Answer 1

The law of sines says that the ratio of the sin of an angle to the length of the side opposite the angle is the same for all three angles in a triangle.

 

So in your case,

 

sin A / a = sin C / c

 

sin 42° / a = sin 96° / 12

.669 / a = .995 / 12

.669 / a = .0829

.669 / .0829 = a

8.07 in = a

 

What about b and B?

 

Since the sum of the angles in a triangle is 180, B must be 180 - (42 + 96) = 42

 

So B is 42°, just as A is.  If two angles in a triangle are equal, the sides opposite them are equal.  Therefore b = a. 

 

So b = 8.07 in.

 

Answer 2

The law is sines is

a/sinA=b/sinB=c/sinC

 

we know that

 

C=96 deg 

c=12 in

A=42 deg

 

B=180-C-A=42

 

since A=B we have an isoceles triangle, and then a=b 

 

a=b=c•sinA/sinC

 

a=b=(12 in)sin42/sin96=8.07