Question

Whe have a triangle ABC, and we now that :
(Sin^2(a)+sin^2(b)-sin^2(c))/(sin(a) sin(b)) =1
Determine the value of c

Answer 1

heh this is a pretty good one

Answer 2

recall the law of sines:$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$$so we have that \(a=k\sin A,b=k\sin B,c=k\sin C\)

Answer 3

now remember the law of cosines:$$c^2=a^2+b^2-2ab\cos C$$plug in for \(a,b,c\):$$k^2\sin^2C=k^2\sin^2A+k^2\sin^2B-2k^2\sin A\sin B\sin C$$

Answer 4

divide by \(k^2\) to get:$$\sin^2A+\sin^2B-2\sin A\sin B\sin C=\sin^2C$$and then subtract \(\sin^2C\) from both sides while adding \(2\sin A\sin B\sin C\):$$\sin^2A+\sin^2B-\sin^2C=2\sin A\sin B\sin C$$

Answer 5

now observe that your equation is equivalent to \(\sin^2A+\sin^2B-\sin^2C=\sin A\sin B\)

Answer 6

oops that \(\sin C\) in \(2\sin A\sin B\sin C\) should be \(\cos C\)

Answer 7

combining the two we get \(\sin A\sin B=2\sin A\sin B\sin C\) i.e.$$\sin (A)\sin(B)(2\cos C-1)=0$$in other words it must be that \(\sin A=0\) or \(\sin B=0\) or that \(2\cos C-1\)

Answer 8

we know that since \(A,B,C\) are arbitrary choices that \(\sin A=0,\sin B=0\) are the same thing reflecting symmetry. we can then focus on merely one i.e. \(\sin A=0\). given that angles in a triangle must lie in \((0,\pi)\) -- which means that \(\sin A=0\) is impossible

Answer 9

we may then focus our attention to \(2\cos C-1=0\) i.e. \(\cos C=1/2\)

Answer 10

the only angle \(0<C<\pi\) such that \(\cos C=1/2\) is \(C=\pi/3\) Q.E.D.

Answer 11

this one took me a long time guys so appreciation would be nice