if sin (A+B) = 1 and sin (A-B)=1/2 . 0<A+B <90°and A>B. then find A and B. plzzz give me the answer.
Answers
Answer:
We know that
Sin (A+B)=1
A+B=90 equation 1
Sin(A-B)=1\2
A-B=30 equation 2
Solving eq.1and2
A+B=90
A=90-B eq.3
Put the value of A in eq.2
90-B-B=30
-2B =30-90
-2B=-60
B=30
Put the value of B in eq.3
A=90-30
A=60
Hence proved
The value of A is 60° and the value of B is 30°.
Given: sin ( A + B ) = 1 and sin ( A - B ) = 1 / 2. 0 < A + B < 90° and A > B.
To Find: The value of A and B.
Solution:
We shall solve the numerical by implementing the following formulas;
sin ( A + B ) = sin A . cos B + cos A . sin B ......(1)
sin ( A - B ) = sin A . cos B - cos A . sin B ......(2)
Coming to the numerical, we have;
sin ( A + B ) = 1 and sin ( A - B ) = 1 / 2
We shall solve these individually;
sin ( A + B ) = 1
⇒ sin A . cos B + cos A . sin B = 1 ......(3)
sin ( A - B ) = 1 / 2
⇒ sin A . cos B - cos A . sin B = 1 / 2 ......(4)
Adding (3) and (4), we get;
⇒ 2 sin A . cos B = ( 1 + 1/2 )
⇒ sin A . cos B = 3 / 4 ......(5)
Putting the value of (5) in (3), we get;
⇒ sin A . cos B + cos A . sin B = 1
⇒ 3 / 4 + cos A . sin B = 1
⇒ cos A . sin B = 1 - ( 3 / 4 )
⇒ cos A . sin B = 1 / 4 .......(6)
Now, it is also said that 0 < A+B < 90°and A > B.
So, we can write (5) as;
sin A . cos B = 3 / 4
⇒ sin A . sin ( 90° - A ) = 3 / 4
⇒ sin A . sin A = 3 / 4
⇒ sin² A = 3 / 4
⇒ sin A = √3 / 2
⇒ sin A = sin 60°
⇒ A = 60°
Since A and B are complementary angles so, we can say that;
A + B = 90°
⇒ 60° + B = 90°
⇒ B = 90° - 60°
= 30°
Hence, the value of A is 60° and the value of B is 30°.
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