In ABC angle B 60 o AB 5 and BC 8 Find the AC to the neare

In ABC, angle B = 60 ^o, AB= 5, and BC = 8.

Find the AC to the nearest tenth.

Find the area of ABC to the nearest tenth

Solution

B = 60 deg ; AB = 5 ; BC = 8

So, angle B is opposite AC .

Apply cosine rule : AC^2 = AB^2 + BC^2 - 2AC*ABcosB

= 5^2 +8^2 -2*5*8cos60

AC = 7 units

Area of traingle : (1/2)*AB*BC*sinB = (1/2)(5*8*sin60)

= 17.32

= 17.3 squnit