Question

if sin(A-B)= 1/2,tan(A+B)=√3, then prove that angle A =45degrees​

Answer 1

sin(A-B)= $\frac{1}{2}$

sin(A-B)= sin 30°

Cancelling sin from both side

A-B = 30° ----> (1)

tan(A+B)=√3

tan(A+B)= tan 60°

Cancelling tan from both side

A+B = 60° ----> (2)

Adding (1) and (2) :

=> 2A = 90°

=> $\fbox{A = 45°}$

Answer 2

Given that,

$\sin(A-B)=\dfrac{1}{2}\\\\\tan(A+B)=\sqrt{3}$

To prove:

$\angle A=45^{\circ}$

Solution,

We have,

$\sin(A-B)=\dfrac{1}{2}$ ......(1)

$\tan(A+B)=\sqrt{3}$ ....(2)

We know that,

$\sin 30=\dfrac{1}{2}$

Equation (1) becomes:

$\sin(A-B)=\sin 30\\\\A-B=30\ .......(3)$

We know that,

$\tan60=\sqrt{3}$

Equation (2) becomes:

$\tan(A+B)=\tan60\\\\A+B=60\ .......(4)$

Solving equation (1) and (2) we get :

$2A=90^{\circ}\\\\A=45^{\circ}$

Hence, proved.

Learn more,

Trigonometry

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