In triangle ABC, side BC is the average of the other two sides. If cos C = AB/AC , find the numerical value of cos C.

Question

Not sure how to solve this since there are no numerical values given. How would you solve this? Law of cosine?

Answer 1

In order to avoid expressions like AC²:

Let a=BC, b = AC, c = AB. (draw a diagram to confirm)

Then a = (b+c)/2 and cosC = c/b.

By the law of cosines:

cos C = (a² + b² - c²) / 2ab

= [(b+c)²/4 + b² - c²] / 2((b+c)/2) (b) -- substitute (b+c)/2 for a

= [(b+c)(b+c)/4]/b(b+c) + (b+c)(b-c)/b(b+c)

= (b+c)/4b + (b-c)/b

= b/4b + c/4b + b/b - c/b

= 1/4 + 1/4 cosC + 1 - cos C --substitute cos C for c/b

cos C = 5/4 -3/4 cos C

7/4 cos C = 5/4

cos C = (5/4) (4/7) = 5/7

Some of the details of the algebraic simplification are left out--hopefully you can fill in the details.

desmos.com/calculator/j452bl0uj5

The above graph depicts different triangles having the given properties. Use the slider for b.

Answer 2

Yes, you would use the law of cosines:

AB is the side opposite angle C. AB^2 = AC^2 + BC^2 - 2*AC*BC*Cos(C)

Substitute AB/AC for cosine C and AB+AC/2 for BC.

Simplify until you get a quadratic equal to 0 so you can evaluate the possible ratios for AB/AC (Cosine C)..