How do you solve the triangle given B=10^circ35', a=40, c=30?

1 Answer
Jan 23, 2017

b= 11.87, C= 27.69^o , A= 141.71^o

Explanation:

Use cosine formula cos B=(a^2 +c^2 -b^2)/(2ac)

Thus cos 10^o 35' = (40^2 +30^2 -b^2)/(2(40)(30))
(for calculation purpose consider 10^o 35'=10.6^o)

0.9829=(2500 -b^2)/2400

b^2= 2500- 0.9829*2400 = 2500-2359=141

b=sqrt 141= 11.87

Next using sine formula sin B/ b=sin C/c

sinC=30/sqrt148sin 10^o 35'=0.4647

C= 27.69^o

A= 180-10^o 35' - 27.69^O = 141.71^o