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Question

$text If end text 0 less than theta less than pi over 2 text and end text 2 sin invisible function application theta equals square root of 3 cos space 10 to the power of ring operator plus sin space 10 to the power of ring operator text then end text theta equals$

$70^{\circ}$
$50^{\circ}$
$60^{\circ}$
$40^{\circ}$

hint Hint:

We will first rearrange the given expression as follows $sin open parentheses theta close parentheses equals fraction numerator square root of 3 over denominator 2 end fraction cos open parentheses 10 to the power of ring operator close parentheses plus 1 half sin open parentheses 10 to the power of ring operator close parentheses$ and then use the formula of
$sin open parentheses A plus B close parentheses$

The correct answer is: $70 to the power of ring operator$

In this question we are given expression
$2 sin invisible function application theta equals square root of 3 cos space 10 to the power of ring operator plus sin space 10 to the power of ring operator$ and we have to find $theta$ .
Step1: Rearranging the equation
We will rearrange the equation by dividing both sides by $2$ . The resulting equation will be
$sin open parentheses theta close parentheses equals fraction numerator square root of 3 over denominator 2 end fraction cos open parentheses 10 to the power of ring operator close parentheses plus 1 half sin open parentheses 10 to the power of ring operator close parentheses$
Step2: Using the formula of $sin open parentheses A plus B close parentheses$
We know that $sin open parentheses A plus B close parentheses space equals space sin open parentheses A close parentheses cos open parentheses B close parentheses plus sin open parentheses B close parentheses cos open parentheses A close parentheses$
We also know that $fraction numerator square root of 3 over denominator 2 end fraction equals sin open parentheses 60 to the power of ring operator close parentheses$ and $1 half equals cos open parentheses 60 to the power of ring operator close parentheses$
Step3: Putting the values in the formula of $sin open parentheses A plus B close parentheses$ we get,
$sin open parentheses 60 to the power of ring operator close parentheses cos open parentheses 10 to the power of ring operator close parentheses plus sin open parentheses 10 to the power of ring operator close parentheses cos open parentheses 60 to the power of ring operator close parentheses$
=> $sin open parentheses 60 to the power of ring operator plus 10 to the power of ring operator close parentheses$
=> $sin open parentheses 70 to the power of ring operator close parentheses$

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