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

Question

2663 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T17:37:06

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

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 #