Question

If sin(A+B)=1 and sin(A-B)=1/2,then the value of A is *

Answer 1

Step-by-step explanation:

sin(a+b)=sin90

a+b=90---1

sin(a-b)=sin30

a-b=30---2

solving 1 and 2

1+2:-

a+b+a-b=90+30

2a=120

a=60

putting value of a in1

b=30

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Answer 2

Given :-

• Sin( A+B ) = 1

• Sin (A-B) = ½

To Find :-

• Value of A

Solution :-

$\sf{\red{\implies} \sin [A+B] = 1 }$

$\sf{\red{\implies} \sin [A-B] = \dfrac{1}{2} }$

$\sf{\red{\implies} \sin [A+B] = \sin \dfrac{\pi}{2} }$

$\sf{\red{\implies} \sin [A-B] = \sin \dfrac{\pi}{6} }$

$\sf{\red{\implies} [A+B] = \dfrac{\pi}{2}\;\; eq1st }$

$\sf{\red{\implies} [A-B] = \dfrac{\pi}{6} \;\; eq2 }$

Adding eq1 from 2nd

$\sf{\red{\implies} A-B + A + B = \dfrac{\pi}{6} + \dfrac{\pi}{2} }$

$\sf{\red{\implies} 2A = \dfrac{\pi}{6} + \dfrac{\pi}{2} }$

$\sf{\red{\implies} 2A = \dfrac{\pi + 3 \pi}{6} }$

$\sf{\red{\implies} 2A = \dfrac{4 \pi}{6} }$

$\sf{\red{\implies} 2A = \dfrac{2 \pi}{3} }$

Taking 2 common both sides

$\sf{\red{\implies}[A] = \dfrac{\pi}{3} }$