If tanθ = 20/21, show that
We have, tanθ = (20k)/(21k) = BC/AC (For some value of k)
By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2
∴AB2 = BC2 + AC2
= AB2 = (20k)2 + (21k)2
= AB2 = 400k2 + 441k2
= AB2 = 841k2
= (29k)2
→ AB = 29k
∴ sinθ = BC/AB = (20k)/(29k) = 20/29
cosθ = AC/AB = (21k)/(29k) = 21/29
consider, the LHS
LHS = =
=
= 30/70
= 3/7
= RHS
HENCE PROVED