Question

If tan(2A+B)=root3 and cot(3A-B)=root3. Find A and B

Answer 1

Answer:

tan(2a+b)=√3

tan(2a+b)= tan60

2a+b=60-----1

cot(3a-b) =√3

cot(3a-b) =cot30

3a-b=30------2

2a+b=60

(+) 3a-b=30

5a=90

a=90/5=18

2a+b=60

2(18)+b=60

b=60-36

b=24

Step-by-step explanation:

BTS ARMY

Answer 2

Answer:-

$\sf{The \ value \ of \ A \ and \ B \ are \ 18 \ and}$

$\sf{24 \ respectively.}$

Given:

• tan(2A+B)=√3

• cot(3A-B)=✓3

To find:

.

• Value of A and B.

Solution:

$\sf{According \ to \ the \ first \ condition.}$

$\sf{tan(2A+B)=\sqrt3}$

$\sf{But, \ we \ know \ tan60^\circ=\sqrt3}$

$\sf{\therefore{2A+B=60...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{cot(3A-B)=\sqrt3}$

$\sf{But, \ we \ know \ tan30^\circ=\sqrt3}$

$\sf{\therefore{3A-B=30...(2)}}$

$\sf{Add \ equations (1) \ and \ (2), \ we \ get}$

$\sf{2A+B=60}$

$\sf{+}$

$\sf{3A-B=30}$

____________________

$\sf{5A=90}$

$\sf{\therefore{A=\frac{90}{5}}}$

$\boxed{\sf{\therefore{A=18}}}$

$\sf{Substitute \ A=18 \ in \ equation (1)}$

$\sf{2(18)+B=60}$

$\sf{\therefore{36+B=60}}$

$\sf{\therefore{B=60-36}}$

$\boxed{\sf{\therefore{B=24}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ A \ and \ B \ are \ 18 \ and}}}$

$\sf\purple{\tt{24 \ respectively.}}$