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)?

1 Answer
Feb 25, 2017

#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#