Answer:
The Law of Cosine : cos C = [tex]\frac{a^{2}+ b^{2}-c^{2}}{2ab}[/tex]
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 = [tex]\frac{BD}{BC}[/tex]
Or, cos C = [tex]\frac{x}{a}[/tex]
∴ x= a cos C
Now put ht value of x in eq 3
I.e, c² = a² + b² - 2ab cos C
Hence , cos C = [tex]\frac{a^{2}+ b^{2}-c^{2}}{2ab}[/tex] Proved Answer
