How do I solve a triangle with side a=8 and b=10 and c=12 using law of sines and cosines?????? any help appreciated and all steps shown out would be super great thanks! :)
+130561 We can use the Law of Cosines to find one angle and then the Law of Sines to find the next one. Choosing any one side, we have
12^2 = 10^2 + 8^2 -2(10)(8)cosC simplify
144 = 100 + 64 - 160cosC which implies that .....
[144 - 100 - 64] / [ -160] = cosC
1/8 = cosC
Using the inverse cosine, we have
cos-1 (1/8) = C = about 82.8°
And using the Law of Sines, we can find B, thusly:
sin B / 10 = sin 82.8 / 12
sinB = 10sin(82.8) / 12
sin-1 [10sin(82.8) / 12] = B = about 55.77°
And angle A can be found by subtracting the sum of B and C from 180
+130561 We can use the Law of Cosines to find one angle and then the Law of Sines to find the next one. Choosing any one side, we have
12^2 = 10^2 + 8^2 -2(10)(8)cosC simplify
144 = 100 + 64 - 160cosC which implies that .....
[144 - 100 - 64] / [ -160] = cosC
1/8 = cosC
Using the inverse cosine, we have
cos-1 (1/8) = C = about 82.8°
And using the Law of Sines, we can find B, thusly:
sin B / 10 = sin 82.8 / 12
sinB = 10sin(82.8) / 12
sin-1 [10sin(82.8) / 12] = B = about 55.77°
And angle A can be found by subtracting the sum of B and C from 180
