in sin θ = 7/25 … [Given]
We know that,
sin2 θ + cos2 θ = 1
…[Taking square root of both sides]
Now, tanθ = sin\(\theta\)/cosθ
Alternate Method:
sin θ = 7/25 …(i) [Given]
Consider ∆ABC, where ∠ABC 90° and ∠ACB = θ.
sin θ = AB/AC … (ii) [By definition]
∴ AB/AC = 7/25 … [From (i) and (ii)]
LetAB = 7k and AC = 25k
In ∆ABC, ∠B = 90°
∴ AB2 + BC2 = AC2 … [Pythagoras theorem]
∴ (7k)2 + BC2 = (25k)2
∴ 49k2 + BC2 = 625k2
∴ BC2 = 625k2 – 49k2
∴ BC2 = 576k2
∴ BC = 24k …[Taking square root of both sides]