If tan(2A B) 3 and cot(3A – B) 3 + = = . Find A and B.
Answers
Answer:
tan(2A+B)=√3
cot(3A-B)=✓3
To find:
.
Value of A and B.
Solution:
\sf{According \ to \ the \ first \ condition.}According to the first condition.
\sf{tan(2A+B)=\sqrt3}tan(2A+B)=
3
\sf{But, \ we \ know \ tan60^\circ=\sqrt3}But, we know tan60
∘
=
3
\sf{\therefore{2A+B=60...(1)}}∴2A+B=60...(1)
\sf{According \ to \ the \ second \ condition.}According to the second condition.
\sf{cot(3A-B)=\sqrt3}cot(3A−B)=
3
\sf{But, \ we \ know \ tan30^\circ=\sqrt3}But, we know tan30
∘
=
3
\sf{\therefore{3A-B=30...(2)}}∴3A−B=30...(2)
\sf{Add \ equations (1) \ and \ (2), \ we \ get}Add equations(1) and (2), we get
\sf{2A+B=60}2A+B=60
\sf{+}+
\sf{3A-B=30}3A−B=30
____________________
\sf{5A=90}5A=90
\sf{\therefore{A=\frac{90}{5}}}∴A=
5
90
\boxed{\sf{\therefore{A=18}}}
∴A=18
\sf{Substitute \ A=18 \ in \ equation (1)}Substitute A=18 in equation(1)
\sf{2(18)+B=60}2(18)+B=60
\sf{\therefore{36+B=60}}∴36+B=60
\sf{\therefore{B=60-36}}∴B=60−36
\boxed{\sf{\therefore{B=24}}}
∴B=24
\sf\purple{\tt{\therefore{The \ value \ of \ A \ and \ B \ are \ 18 \ and}}}∴The value of A and B are 18 and
\sf\purple{\tt{24 \ respectively.}}24 respectively.