Answer
A=$s^{2}$cos θ sinθ
Work Step by Step
Lets take triangle ABC as isosceles triangle. AB =s, AC=s
AD is the altitude so it divides BC in two BD and DC
we know BD = DC = $\frac{base}{2}$
Lets take BD = DC = x units
in triangle ADB
cos θ = $\frac{AD}{AB}$ = $\frac{AD}{s}$
AD = (cos θ)s
in triangle ADC
sin θ = $\frac{DC}{AC}$ = $\frac{x}{s}$
x = (sin θ)s
Area of triangle= $\frac{1}{2}$bh
=$\frac{1}{2}$*BC * AD
=$\frac{1}{2}$* 2*s(cos θ)*s(sinθ)
A=$s^{2}$cos θ sinθ
