# Math Help - Can someone find the mistake?

1. ## Can someone find the mistake?

Solve the triangle:
b=2 c=3 A=95 (Capital letters are angles, lower case is sides.)

a=sqrt(2^2+3^2-2(2)(3)cos(95))
a= 3.7 (Rounded.)

b^2=a^2+c^2-2(a)(c)cosB
2(a)(c)cosB=a^2+c^2-b^2
cosB=[a^2+c^2-b^2]/[2(a)(c)]
B=arccos([a^2+c^2-b^2]/[2(a)(c)])
B=arccos([3.7^2+3^2-2^2]/[2(3.7)(3)])=32.66 (Rounded)

By the same logic then...
C=arccos([3.7^2+4^2-3^2]/[2(3.7)(2)])=54.04

However;
95+32.66+54.04=181.7
Ergo, something's wrong, but what? I think B might equal 30.96, but I don't know where I went wrong in the math...

2. ## Re: Can someone find the mistake?

this error happened from the rounding process it is normal nothing wrong

3. ## Re: Can someone find the mistake?

You have a Typo in your expression for B, but your calculation used the correct value.

I'm pretty sure that the reason for your angle discrepancy is that you rounded the value for a quite severely. The actual value is more like 3.748 .

That's more than a 1% difference. Your angle error is about 1% also.