If A(0 ,0), B(θ , cos θ ) and C((sin)3 θ , 0) are the vertices of a triangle A B C, then the value of θ for which the triangle has the maximum area is (where .θ ∈(0, (π/2)))

Question

If A(0 ,0), B(θ , cos θ ) and C((sin)3 θ , 0) are the vertices of a triangle A B C, then the value of θ for which the triangle has the maximum area is (where .θ ∈(0, (π/2))) (A) (π /6) (B) (π /4) (C) (π /3) (D) (π /2)

Tardigrade Admin · Accepted Answer

Area of triangle ABC, triangle =| (1/2) | 0 0 1 θ cos θ 1 sin3â¡θ 0 1 | | ⇒ triangle =(1/2)sin3 θ cosâ¡θ (d triangle /d θ )=(3 sin2 θ cos2 â¡ θ - sin4 â¡ θ /2)=0 ⇒ (sin)2 θ (3 (cos)2 â¡ θ - (sin)2 â¡ θ )=0 ⇒ sin θ =0 or tan2 θ =3 θ =0 or θ =(π /3) ((d2 triangle /d (θ )2)) < 0forθ =(π /3)

Q. If $A (0, 0), B (θ, cos θ)$ and $C ((s in)^{3} θ, 0)$ are the vertices of a triangle $A BC,$ then the value of $θ$ for which the triangle has the maximum area is (where $θ \in (0, \frac{π}{2}))$

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

Q. If A(0,0),B(θ,cosθ) and C((sin)3θ,0) are the vertices of a triangle ABC, then the value of θ for which the triangle has the maximum area is (where θ∈(0,2π​))

6π​

4π​

3π​

2π​

Area of triangle ABC, △=∣∣​21​∣∣​0θsin3⁡θ​0cosθ0​111​∣∣​​∣∣​ ⇒△=21​sin3θcos⁡θ dθd△​=23sin2θcos2⁡θ−sin4⁡θ​=0 ⇒(sin)2θ(3(cos)2⁡θ−(sin)2⁡θ)=0 ⇒sinθ=0 or tan2θ=3 θ=0 or θ=3π​ (d(θ)2d2△​)<0forθ=3π​

Q. If $A (0, 0), B (θ, cos θ)$ and $C ((s in)^{3} θ, 0)$ are the vertices of a triangle $A BC,$ then the value of $θ$ for which the triangle has the maximum area is (where $θ \in (0, \frac{π}{2}))$

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$