Solve triangle ABC.

Question

b = 30,c= 60, ∠A = 70^°

Answer 1

Greetings! Lets solve this shall we ?

So, we must use the Law of Cosines to complete the triangle such that

a² = b² + c² - 2bccosA since we were given b, c and Angle A. Then,

a² = (30)² + (60)² - 2(30)(60)cos(70)

a² = 900 + 3600 - 3600cos70

a² = 4500 - 3600cos70

a = √(3268.727484) = 57.17278622 = 57.173

Now, we use the Law of Sines to find Angle B and Angle C such that

sinA/a = sinB/b

sin70/(57.173) = sinB/30........Cross multiply to get

30sin70 = 57.173sinB

sinB = (30sin70) / (57.173)

Angle B = sin^-1(30sin70/57.173) = 29.54^o

Angle C = 180 - (70+29.54) = 180 - 99.543 = 80.46^o

I hope this helped!