Question

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Answer 1

Answer:

....................

Answer 2

SOLUTION :

sin ( A - B ) = 1/2

=> sin ( A - B ) = sin 30

=> A - B = 30 _________ ( 1 ) eqn

cos ( A + B ) = 1/2

=> cos ( A + B ) = cos 60

=> A + B = 60 ___________ ( 2 ) eqn

adding these two eqns

( A + B ) + ( A - B ) = 30 + 60

=> 2 A = 90

=> A = 45 [ Ans ]

substituting the value of A in eqn ( 1 )

A - B = 30

=> 45 - B = 30

=> B = 15 [ Ans ]

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learn more :

$\checkmark \:\: \sf \sin ( A + B ) = \sin A \cos B + \cos A \sin B$

$\checkmark \:\: \sf \sin ( A - B ) = \sin A \cos B - \cos A \sin B$

$\checkmark \:\: \sf \cos ( A + B ) = \cos A \cos B - \sin A \sin B$

$\checkmark \:\: \sf \cos ( A - B ) = \cos A \cos B + \sin A \sin B$

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table

$\begin{array}{c|c|c|c|c|c|} \sf ratios &\underline{0\degree}&\underline{30\degree}&\underline{45\degree}&\underline{60\degree}&\underline{90\degree}\\ \sin & 0 &\frac{1}{2}&\frac{1}{\sqrt2}&\frac{\sqrt3}{2}&1\\\cos & 1 & \frac{\sqrt3}{2}&\frac{1}{\sqrt2}&\frac{1}{2}&0\end{array}$

lets find out another ratios like tan, cot, sec, cosec

$\begin{array}{c|c|c|c|c|c|} \sf ratios &\underline{0\degree}&\underline{30\degree}&\underline{45\degree}&\underline{60\degree}&\underline{90\degree}\\ \frac{ \sin}{\cos}=\tan & \frac{0}{1}=0 &\frac{\frac{1}{2}}{\frac{\sqrt3}{2}}=\frac{1}{\sqrt3}&\frac{\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=1 &\frac{\frac{\sqrt3}{2}}{\frac{1}{2}}=\sqrt3&\frac{1}{0}= meaningless \\ \frac{1}{\tan}=\cot &meaningless & {(\frac{\sqrt3}{2})}^{-1}=\frac{2}{\sqrt3}&{(\frac{1}{\sqrt2})}^{-1}=\sqrt2&{(\frac{1}{2})}^{-1}=2& 0 \end{array}$

try to find out another ratios ( sec , cosec )

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Ans for 10th std student

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