Question

If$0<\theta <\mathrm{\pi },0<\varnothing <\mathrm{\pi }\&\mathrm{cos\theta }.\cos \varnothing .\cos (\mathrm{\theta }+\varnothing )=\frac{-1}{8}$, then

• A. $\theta =\frac{\mathrm{\pi }}{4},\varnothing =\frac{\mathrm{\pi }}{3}$
• B. $\theta =\frac{\mathrm{\pi }}{3},\varnothing =\frac{\mathrm{\pi }}{4}$
• C. $\theta =\varnothing =\frac{\mathrm{\pi }}{3}$
• D. $\theta =\varnothing =\frac{\mathrm{\pi }}{4}$

Answer 1

The correct option is C

$\theta =\varnothing =\frac{\mathrm{\pi }}{3}$

Finding the value:

By comparing each term to the numerical value.

We have given$0<\theta <\mathrm{\pi },0<\varnothing <\mathrm{\pi }\&\mathrm{cos\theta }.\cos \varnothing .\cos (\mathrm{\theta }+\varnothing )=\frac{-1}{8}$,

$\mathrm{cos\theta }.\cos \varnothing .\cos (\mathrm{\theta }+\varnothing )={\left(\frac{-1}{2}\right)}^3$…….$\left(1\right)$

from$\left(1\right)$,

we can take$\cos \theta =\cos \varnothing =\cos (\theta +\varnothing )=-\frac{1}{2}$

from the above, $\theta =\varnothing =\frac{2\mathrm{\pi }}{3}$ this is not present in the option.

another possibility is,

$\mathrm{cos\theta }.\cos \varnothing .\cos (\mathrm{\theta }+\varnothing )=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)$…….$\left(2\right)$

from$\left(2\right)$,

we can take$\begin{array}{rcl}\cos \theta & =& \cos \varnothing =\frac{1}{2},\cos (\theta +\varnothing )=-\frac{1}{2}\end{array}$ all these values are satisfying the given conditions,

from the above, $\theta =\varnothing =\frac{\mathrm{\pi }}{3}\&\theta +\varnothing =\frac{2\mathrm{\pi }}{3}$

Hence, the option$\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is the correct option.