Question

How do you solve the triangle when sides of a = 6, b = 10 and measure of angle A = (in degrees, minutes, seconds) 31˚10', find the other sides (c) and other angle measures (B)(C)?

Answer 1

`c=11.593`, `/_B=59.695^@` and `/_C=89.228^@`

Explanation:

To find the third side given two other sides `a=6` and `b=10` and included angle `/_A=31^@10'=31.167^@`, as `10'` is one-sixth of `1^@`). Now using sine formula

`a/sinA=b/sinB=c/sinC` i.e.

`6/(sin31.167^@)=10/sinB=c/sinC`

or `6/0.5175=10/sinB=c/sinC`

Hence `sinB=(10xx0.5175)/6=0.8626` and `/_B=59.695^@`

and `/_C=180^@-31.167^@-59.695^@=89.228^@`

and `c=6/0.5175xxsin89.228^@=6/0.5175xx0.9999=11.593`