Question

If sin(A-B)=1/2,cos(A+B)=1/2,0°A+BB,find A and B​

Answer 1

Gɪᴠᴇɴ :-

$\sf \bull \: sin ( A - B ) = \dfrac{1}{2} = 30^\circ$

$\sf \bull \: cos(A + B) = \dfrac{1}{2} = 60^\circ$

⠀

ᴛᴏ Fɪɴᴅ :-

$\bull$ The value of A and B

⠀

sᴏʟᴜᴛɪᴏɴ :-

$\sf\bull \: A-B = 30^\circ \: \: \: \: - - \: Eq.1 \\ \\ \sf\bull \: A + B = 60^\circ \: \: \: \: - - \: Eq.2$

Now, we will substitute Eq. 1 and Eq. 2, and solve this by föllowing the method of substitution method. We get,

$\sf : \implies \: A - B + A + B = 30^\circ + 60^\circ \\ \\ \sf : \implies \:2A = 90^\circ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf : \implies \:A = \cancel \dfrac{90}{2} = 45^\circ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now, we will substitute the value of A in equation 2 and get the value of B. We get,

$\sf : \implies \: A + B = 60 ^\circ \: \\ \\ \sf : \implies45 ^\circ + B = 60 ^\circ \: \\ \\ \sf : \implies \: B = 60^\circ - 45^\circ \\ \\ \sf : \implies \:B = 15 ^\circ \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore, the value of A is 45° and B is 15°.

Answer 2

GIVEN :–

$\\\bf(i) \: \: \sin(A-B) = \dfrac{1}{2}$

$\\\bf(ii) \: \: \cos(A + B) = \dfrac{1}{2}$

TO FIND :–

• Measures of Angle A & B .

SOLUTION :–

• According to the first condition –

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

$\\\bf\implies\sin(A-B) = \sin( {30}^{ \circ} )$

$\\\bf\implies A-B={30}^{ \circ} \: \: \: \: \: \: \: \: - - - eq.(1)$

• According to second condition –

$\\\bf\implies\cos(A + B) = \dfrac{1}{2}$

$\\\bf\implies\cos(A + B) = \cos( {60}^{ \circ})$

$\\\bf\implies A+B= {60}^{ \circ} \: \: \: \: \: \: \: - - - eq.(2)$

• Adding eq.(1) & eq.(2) –

$\\\bf\implies A-B + A+B= {60}^{ \circ} + {30}^{ \circ}$

$\\\bf\implies 2A= {90}^{ \circ}$

$\\\large\implies \boxed{ \red{ \bf A= {45}^{ \circ}}}$

• By eq.(1) –

$\\\bf\implies A-B={30}^{ \circ}$

$\\\bf\implies {45}^{\circ}-B={30}^{ \circ}$

$\\\large\implies \boxed{ \red{ \bf B= {15}^{ \circ}}}$

• Henceforth , A is 45⁰ and B is 15⁰.

• Learn more :–

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$