Triangle and angle help,?

Question

Given that ...
a=6
b=4
C=96 degrees

How do I figure out angle A??

Thanks for any help..

Answer 1

96° - 90° = 6°
tanA = 6cos6/(4 + 6sin6)
tanA = 5.9671/4.6273
tanA = 1.2896
A = 52.208°

or
tanA = 6sin96/(4 - 6cos96)
tanA = 5.9671/4.6273

Answer 2

Use the law of cosines:
c^2 = a^2 + b^2 - 2ac cos C

that will give you the value of side c.

then use the law of sines to find A.

a/sin A = c / sin C or sin A/ a = sin C / c
sin A = (a sin C )/ c
A = arcsin [(asinC)/c]

try that and see what you get.

Answer 3

COSINE RULE
c^2 = a^2 + b^2 - 2ab CosC
= 6^2 + 4^2 - 2(6)(4) Cos96
= 36 + 16 - 48 x (-0.1045)
= 36 + 16 + 5.017
= 57.017
c = 7.55

SINE RULE
(Sin A) / a = (Sin B) / b = (Sin C)/c
(Sin A) / 6 = (Sin B) / 4 = (Sin 96)/7.55 = 0.1318
(Sin A) / 6 = 0.1318 >>[(Sin B) / 4 = 0.1318 if needed]<<
Please use the calculator and sail

Answer 4

First use the law of cosines and solve for side C.
c^2 = a^2 +b^2 -2abcos(C)
Once you have side c, you can use the law of sines to find angle A.
sinA / a = sinC / c
then solve for angle A.