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Question

Mathematics Question on Trigonometric Functions

If θ\theta and ϕ\phi are acute angles, sinθ=12,cosϕ=13,\sin \, \theta = \frac{1}{2} , \cos \, \phi = \frac{1}{3}, then the value of θ+ϕ\theta + \phi lies in

A

(π3,π2]\left( \frac{\pi}{3}, \frac{\pi}{2} \right]

B

(π2,2π3]\left( \frac{\pi}{2}, \frac{2\pi}{3} \right]

C

(2π3,5π6]\left( \frac{2\pi}{3}, \frac{5\pi}{6} \right]

D

(5π6,π]\left( \frac{5\pi}{6},\pi \right]

Answer

(π2,2π3]\left( \frac{\pi}{2}, \frac{2\pi}{3} \right]

Explanation

Solution

Now sinθ=12θ=π6\sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} and cosϕ=13π3<ϕ<π2\cos \phi = \frac{1}{3} \Rightarrow \frac{\pi}{3} < \phi < \frac{\pi}{2} (12>13>0,i.e.,cosπ3>cosϕ>cosπ2)\left(\because \frac{1}{2} > \frac{1}{3} > 0 , i.e., \cos \frac{\pi}{3} > \cos \phi > \cos \frac{\pi}{2}\right) Hence π6+π3<θ+ϕ<π6+π2\frac{\pi}{6} + \frac{\pi}{3} < \theta + \phi< \frac{\pi}{6} + \frac{\pi}{2}