Question

5.
If tan(A + B) = 1, and cos(A - B) =
$\sqrt{3 \div 2}$

find A and B.

​

Answer 1

Given :

Values:

•$\rm{ tan(A+B)=1 }$

•$\rm{ cos(A-B) =\dfrac{\sqrt{3}}{2}}$

To Find :

• The value of A and B =?

Solution :

Given :

$\\ \bullet\rm{tan(A+B)=1}$

$\\ \bullet\rm{cos(A-B)=\dfrac{\sqrt{3}}{2}}$

• 0° <A+B < 90° and A > B

We know that,

$\\ \bullet \rm \: tan 45 \degree = 1$

$\\ \bullet \rm \: cos \: 30 \degree = \frac{ \sqrt{3} }{2}$

$\\ \implies\sf{tan(A+B)=1 }$

$\\ \implies\sf{A+B=45 \degree}$

•A + B = 45° ............(1)

$\\ \implies\sf{cos(A-B)=\dfrac{\sqrt{3}}{2}}$

$\\ \implies\sf{A-B=30 \degree}$

•A - B = 30 ............(2)

Adding Equations (1) and (2) :

$\bullet\bf{A+\cancel{B}=45 \degree}$

$\\ \bullet\bf{A- \cancel{B}=30 \degree}$

_______________________________

$\\ \implies\sf{2A=75 \degree}$

$\\ \implies\sf{A=\dfrac{75}{2}}$

$\\ \implies\sf{A=37.5 \degree}$

Putting the value of A in equation (1) we get,

$\\ \implies\sf{37.5 \degree+B=45 \degree}$

$\\ \implies\sf{B=45 - 37.5}$

$\\ \implies\sf{B=7.5 \degree}$

Therefore,

• The value of A is 37.5°

• The value of B is 7.5°

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{tan(A + B) = 1} \\ &\sf{cos(A - B) = \dfrac{ \sqrt{3} }{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find -   \begin{cases} &\sf{A \: and \: B}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

★ Given,

$:\implies \sf \: tanA \: = \: 1$

$:\implies \sf \: tanA \: = \: tan45 \degree$

$:\implies \sf \: A + B = 45 \degree - - - (1)$

★ Again,

$:\implies \sf \: cos(A - B) = \dfrac{ \sqrt{3} }{2}$

$:\implies \sf \: cos(A - B) = cos30 \degree$

$:\implies \sf \: A - B = 30 \degree - - - (2)$

★ On adding equation (1) and (2), we get

$:\implies \sf \: A + B + A - B = 45 \degree + 30\degree$

$:\implies \sf \: 2A = 75 \degree$

$:\implies \sf \: A \: = \: 37.5 \degree$

★ Now, substituting the value of A in equation (1), we get

$:\implies \sf \: 37.5\degree \: + \: B = 45\degree$

$:\implies \sf \: B \: = \: 7.5\degree$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\bf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$