Solve the triangles with the given parts using Law of Cosines:

Question

Solve the triangles with the given parts using Law of Cosines:

find : m∠C =?  °, m∠B =  ?°, BC = ?

  

Answer 1

The Law of Cosines states c² = a² + b² - 2(a)(b)cosC

From the diagram, BC² = 17² + 26² - 2(17)(26)cos(88)

So BC² = 289 + 676 - 884(0.0349) = 934

So BC = √934 = 30.56

From here it's very easy to use the Law of Sines to find the remaining angles but your instructions are to use the Law of Cosines.

If we denote the lengths opposite the angles B and C as b and c we can write out the equations in an easier to understand format. Here we can let BC = a = 30.56 since the opposite angle is angle A, b = 26, and c = 17

c² = a² + b² - 2abcosC

17² = (30.56)² + 26² - 2(30.56)(26)cosC

Solve for cosC then for C

26² = 17² + (30.56)² - 2(17)(30.56)cosB

Solve for cosB then for B