Question

this is the question ​

Answer 1

$\large\underline{\underline\text{Question:}}$

• In Attachment.

.

$\large\underline{\underline\text{Solution:}}$

As given,

• $\sin( \theta - \phi) = \frac{1}{2}$

As we know that,

• $\sin( {30}^{ \circ} ) = \frac{1}{2}$

So we can say that,

• $\sin( \theta - \phi) = \sin( {30}^{ \circ} )$

Comparing both sides,

• $\theta - \phi = {30}^{ \circ} \: \: \: \: \: \: ...(1)$

As given,

• $\cos( \theta + \phi) = \frac{1}{2}$

As we know that,

• $\cos( {60}^{ \circ} ) = \frac{1}{2}$

So we can say that,

• $\cos( \theta + \phi) = \cos( {60}^{ \circ} )$

Comparing both sides,

• $\theta + \phi = {60}^{ \circ} \: \: \: \: \: \: ...(2)$

By Elimination method we can find the values,

We have,

• $\theta - \phi = {30}^{ \circ} \: \: \: \: \: \: ...(1)$

• $\theta + \phi = {60}^{ \circ} \: \: \: \: \: \: ...(2)$

Let substract Eq [1] from Eq [2],

$\cdot \cdot \cdot \longrightarrow ( \theta + \phi) - ( \theta - \phi) = {60}^{ \circ} - {30}^{ \circ}$

Opening the brackets,

$\cdot \cdot \cdot \longrightarrow \theta + \phi- \theta + \phi = {60}^{ \circ} - {30}^{ \circ}$

$\cdot \cdot \cdot \longrightarrow 2 \phi = {30}^{ \circ}$

$\cdot \cdot \cdot \longrightarrow \phi = \frac{{30}^{ \circ}}{2}$

We get,

$\cdot \cdot \cdot \longrightarrow \boxed{\phi = {15}^{ \circ} }$

$\large\underline{\underline\text{Required Answer:}}$

• Option B) 15°.