If θ and φ are acute angles, sin θ = (1/2) , cos φ = (1/3), then the value of θ + φ lies in

Question

If θ and φ are acute angles, sin θ = (1/2) , cos φ = (1/3), then the value of θ + φ lies in (A) ( (π/3), (π/2) ] (B) ( (π/2), (2π/3) ] (C) ( (2π/3), (5π/6) ] (D) ( (5π/6),π ]

Tardigrade Admin · Accepted Answer

Now sinθ = (1/2) ⇒ θ = (π/6) and cos φ = (1/3) ⇒ (π/3) < φ < (π/2) (∵ (1/2) > (1/3) > 0 , i.e., cos (π/3) > cos φ > cos (π/2)) Hence (π/6) + (π/3) < θ + φ< (π/6) + (π/2)

Q. If $θ$ and $ϕ$ are acute angles, $sin θ = \frac{1}{2}, cos ϕ = \frac{1}{3},$ then the value of $θ + ϕ$ lies in

$(\frac{π}{3}, \frac{π}{2}]$

$(\frac{π}{2}, \frac{2 π}{3}]$

$(\frac{2 π}{3}, \frac{5 π}{6}]$

$(\frac{5 π}{6}, π]$

Now $sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$
and $cos ϕ = \frac{1}{3} \Rightarrow \frac{π}{3} < ϕ < \frac{π}{2}$
$(∵ \frac{1}{2} > \frac{1}{3} > 0, i . e ., cos \frac{π}{3} > cos ϕ > cos \frac{π}{2})$
Hence $\frac{π}{6} + \frac{π}{3} < θ + ϕ < \frac{π}{6} + \frac{π}{2}$

Q. If θ and ϕ are acute angles, sinθ=21​,cosϕ=31​, then the value of θ+ϕ lies in

(3π​,2π​]

(2π​,32π​]

(32π​,65π​]

(65π​,π]