Trig help?

Question

I have a triangle, side a-b is 8cm. Side a-c is 14 cm. Angle c is 32 Degrees. I need to find Angle b. Please tell me how you worked this out?

its not a right angle triangle

Answer 1

If it is a right angled triangle, then the internal angles add up to 180 degrees. Therefore 180 degrees minus 122 degrees ( 90+32 ) is 58 degrees.
Otherwise, is it the opposite (ab) over the adjacent (ac) i.e the Tan of 8/14 would give you angle A. Then subtract angle A+C from 180 degrees to give angle b????
I'm not sure, becuase it's over 10 years since I dig trigonometry at college, and do you know what? I've never had call to use it since! Tell that to your teacher.

Answer 2

It really depends. Is it a right triangle, or a normal triangle?

If it's a right triangle, then one angle is obviously 90 degrees, and the other would be 58 degrees.
If it's a normal triangle, then the other two angles should be 29 degrees.

But don't take my word for it. I'm only in Algebra 2, but I learned some trig in Geometry last year.
Hope I could have helped a little bit <=].

Answer 3

Triangle Angle B at the top? A bottom left&C to the right?
SinB=bxSinA...divided by side a

Answer 4

Remember

T O A - S O H - C A H ?

Tan 32 =0.625

Cos 32 = 0.848

Sin 32 = 0.53

Substitute in correct expression:

Hope this helps :-)

Answer 5

14 / sin B = 8 / sin 32°
14 sin 32° = 8 sin B
sin B = 14 sin 32° / 8
sin B = 0.927
B = 68°

Answer 6

Use the formula a/Sin A = b/Sin B = c/Sin C where 'a' is the side opposite the angle A, 'b' is the side opposite the angle B, and 'c' is the side opposite the angle C.

Answer 7

You can use sine law: a/sinA = b/sinB

8/sin32 = 14/sinB

(14*sin32)/8 = sinB
sinB = 0.927358712...
B = 68.03 ˚