Question

plz help me out of these​

Answer 1

Answer:

It is given that,

tan(2A+3B)=√3

tan(2A+3B)=tan60°

2A+3B=60°→1

tan(3A+2B)=1/√3

tan(3A+2B)=tan30°

3A+2B=30°→2

Solve Equation 1 and 2

B=24

Substitute B value in equation 1

2A+3(24)=60

2A+72=60

2A=-12

A=-6

So, The value of A is -6 and B is 24.....

Step-by-step explanation:

Hope it helps you........

Answer 2

QUESTION :–

• If tan(2A + 3B) = √3 and tan(3A + 2B) = 1/√3 , then find A and B .

ANSWER :–

GIVEN :–

▪︎ tan(2A + 3B) = √3 and tan(3A + 2B) = 1/√3.

TO FIND :–

• A & B = ?

SOLUTION :–

• Let's solve –

=> tan(2A + 3B) = √3

• We should write this as –

=> tan(2A + 3B) = tan(60°)

=> 2A + 3B = 60° ______________eq.(1)

• Second condition –

=> tan(3A + 2B) = 1/√3

=> tan(3A + 2B) = tan(30°)

=> 3A + 2B = 30° ______________eq.(2)

• Multiplying the eq.(1) by '2' –

=> 4A + 6B = 120° ______________eq.(3)

• And Multiplying the eq.(2) by '3' –

=> 9A + 6B = 90° ______________eq.(4)

• Subtract eq.(3) by eq.(4) –

=> 9A + 6B - 4A - 6B = 90° - 120°

=> 5A = -30°

=> A = - 6°

• Using eq.(1) –

=> 2A + 3B = 60°

=> 2(-6°) + 3B = 60°

=> 3B - 12° = 60°

=> 3B = 72°

=> B = 24°