If sin ( A - B ) = frac{1}{2},cos ( A + B ) = | vedclass

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Question

(A) Given that $\sin ( A - B ) = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $A - B = 30^{\circ}$ ......$(1)$Also,given that $\cos ( A

Vedclass · Accepted Answer

$A = 45^{\circ}, B = 15^{\circ}$

Vedclass · Answer

(A) Given that $\sin ( A - B ) = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $A - B = 30^{\circ}$ ......$(1)$Also,given that $\cos ( A + B ) = \frac{1}{2}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $A + B = 60^{\circ}$ ......$(2)$Adding equations $(1)$ and $(2)$:$(A - B) + (A + B) = 30^{\circ} + 60^{\circ}$$2A = 90^{\circ}$$A = 45^{\circ}$Substituting $A = 45^{\circ}$ in equation $(2)$:$45^{\circ} + B = 60^{\circ}$$B = 60^{\circ} - 45^{\circ}$$B = 15^{\circ}$Thus,$A = 45^{\circ}$ and $B = 15^{\circ}$.

If $\sin ( A - B ) = \frac{1}{2}$,$\cos ( A + B ) = \frac{1}{2}$,$0^{\circ} < A + B \leq 90^{\circ}$,and $A > B$,find $A$ and $B$.

Similar Questions

In a right triangle $ABC$,right-angled at $B$. If $\tan A = 1$,then verify that $2 \sin A \cos A = 1$.

State whether the following is true or false. Justify your answer.
$\sin (A+B) = \sin A + \sin B$

Prove the following identities,where the angles involved are acute angles for which the expressions are defined:
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

In $\triangle OPQ$,right-angled at $P$,$OP = 7\, cm$ and $OQ - PQ = 1\, cm$. Determine the values of $\sin Q$ and $\cos Q$.

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

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