Question

If sin square A = 1/2 tan square 45° where A is an acute angle then find the value of A ​

Answer 1

answer is 15 degree banasthali school

Answer 2

Answer :

Given : sin square A = 1/2 tan square 45°

To find : The value of A

Solution : sin²A = $\dfrac{1}{2}$ tan²45°

=> Sin²A = $\dfrac{1}{2}$ × (1)²

Note : Tan 45° = 1

=> sin²A = $\dfrac{1}{2}$

=> sin A = $\dfrac{\sqrt{1}}{\sqrt{2}}$

=> sin A = $\dfrac{1}{\sqrt{2}}$

Note : Sin is 1/√2 at 45°

=> A = 45°