UPSEE Solved Model Paper 5

In ΔACB,
b cos θ = a ... (i)
Given b cos θ + b = 4 (Since b + a = 4 given) b =

4

1+cosθ

From eq (i) a =

4cosθ

1+cosθ

∴ Area of triangle Δ =

1

2

basinθ =

1

2

.

4

1+cosθ

.

4cosθ

1−cosθ

sinθ
=

4sin2θ

(1+cosθ)2

∴

dΔ

dθ

=
=

8[cos2θ(1+cosθ)+sin2θsinθ]

(1+cosθ)3

For maxima or minima
Put

dΔ

dθ

= 0
⇒ cos 2θ (1 + cos θ) + sin 2θ sin θ = 0
⇒ cos 2θ + (cos 2θ cos θ + sin 2θ sin θ) = 0
⇒ cos 2θ + cos θ = 0
∴ cos 2θ = - cos θ = cos (π - θ) ⇒ θ =

π

3

∴ Δ is maximum when θ =

π

3