Question

If cos(81°+theta)=sin(k/3-theta) then find the value of k

Answer 1

the answer is completed

Answer 2

Answer:

Required value of k is 27°.

Step-by-step explanation:

Here given

$\cos( {81}^{o} + \theta ) = \sin( \frac{k}{3} - \theta) ....(1)$

Here we want to find value of k.

For solving this problem first we should know relation between $\sin(\theta) \: \: and \: \: \cos(\theta)$

I.e,$\cos(\theta) = \sin( {90}^{o} -\theta )$

From equation (1),

$\sin( {90}^{o} - ( {81}^{o} + \theta ) = \sin( \frac{k}{3} -\theta ) \\ \sin( {90}^{o} - {81}^{o} - \theta) = \sin( \frac{k}{3} -\theta ) \\ {90}^{o} - {81}^{o} - \theta = \frac{k}{3} -\theta$

${90}^{o} - {81}^{o} = \frac{k}{3} \\ \frac{k}{3} = {9}^{o} \\ k = {9}^{o} \times 3 \\k = {27}^{o}$

Required value of k is 27°.

Some extra important formula of this chapter,

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$