Question

Solve the Law of Cosine: c^2 = a^2+ b^2 - 2abcosC for cos C.

Answer 1

Answer:

The Law of Cosine :  cos C = $\frac{a^{2}+ b^{2}-c^{2}}{2ab}$

Step-by-step explanation:

See the figure to understand the proof :

Let A Triangle ABC with sides a,b,c,

Draw a perpendicular on base AC of height H meet at point D

Divide base length b as AD = x -b   and    CD = x

By Pythagoras Theorem

In Triangle BDC             And     In Triangle BDA

a² = h² + x²     (  1  )                        c² = h² + (x-b)²

                                                      c² = h² + x² + b² - 2xb   ...(. 2)

From above eq 1 and 2

c² = (a² - x²) + x² + b² - 2xb

or, c² = a² + b² - 2xb                    .....(3)

Again in ΔBDC

cos C = $\frac{BD}{BC}$

Or, cos C = $\frac{x}{a}$

∴ x= a cos C

Now put ht value of x in eq 3

I.e, c² = a² + b² - 2ab cos C

Hence , cos C = $\frac{a^{2}+ b^{2}-c^{2}}{2ab}$      Proved   Answer