Derive Law of Cosines using Pythagorean Theorem

Consider the triangle ABC. Denote h the altitude through B and D the point where h intersects the (extended) base AC
Cosine function for triangle ADB.

cos α= x/c  =>  x=c*cos α
 

Pythagorean theorem for triangle ADB
x2+h2=c2*x2+h2=c2
h2=c2−x2*h2=c2−x2

Pythagorean theorem for triangle CDB
(b−x)2+h2=a2*(b−x)2+h2=a2

Substitute h2 = c2 - x2
(b−x)2+(c2−x2)=a2(b−x)2+(c2−x2)=a2
(b2−2bx+x2)+(c2−x2)=a2(b2−2bx+x2)+(c2−x2)=a2
b2−2
bx+c2=a2b2−2bx+c2=a2

Substitute x = ccos α
b2−2b
(ccosα)+c2=a2b2−2b(c*cos α)+c2=a2

Rearrange to get Law of Cosines

a2=b2+c2−2bc*cos α

Answered by Jan M. Maths tutor

2569 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Sketch the graph of y=3sin(2x +pi/2)


I've been told that I can't, in general, differentiate functions involving absolute values (e.g. f(x) = |x|). Why is that?


Find the equation of the straight line perpendicular to 3x+5y+6=0 that passes through (3,4)


show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences