Given: \( \sin \mathrm{P}+\sin \mathrm{Q}=\frac{7}{5} \) and \( \hat{\mathrm{P}}+\hat{\mathrm{Q}}=90^{\circ} \) Without using a calculator, determine the value of \( \sin 2 \mathrm{P} \).

Given that \( \sin \mathrm{P} + \sin \mathrm{Q} = \frac{7}{5} \) and \( \hat{\mathrm{P}} + \hat{\mathrm{Q}} = 90^{\circ} \), we can use the fact that \( \hat{\mathrm{Q}} = 90^{\circ} - \hat{\mathrm{P}} \). This implies that \( \sin \mathrm{Q} = \cos \mathrm{P} \). Now we can rewrite the first equation: \[ \sin \mathrm{P} + \cos \mathrm{P} = \frac{7}{5} \] We will square both sides to make use of the identity \( \sin^2 \mathrm{P} + \cos^2 \mathrm{P} = 1 \): \[ (\sin \mathrm{P} + \cos \mathrm{P})^2 = \left(\frac{7}{5}\right)^2 \] Expanding the left hand side: \[ \sin^2 \mathrm{P} + 2\sin \mathrm{P} \cos \mathrm{P} + \cos^2 \mathrm{P} = \frac{49}{25} \] Substituting \( \sin^2 \mathrm{P} + \cos^2 \mathrm{P} = 1 \): \[ 1 + 2\sin \mathrm{P} \cos \mathrm{P} = \frac{49}{25} \] Subtracting 1 from both sides gives: \[ 2\sin \mathrm{P} \cos \mathrm{P} = \frac{49}{25} - 1 = \frac{49}{25} - \frac{25}{25} = \frac{24}{25} \] We know that \( \sin 2\mathrm{P} = 2\sin \mathrm{P} \cos \mathrm{P} \). Therefore, \[ \sin 2\mathrm{P} = \frac{24}{25} \] Thus, the value of \( \sin 2 \mathrm{P} \) is \[ \boxed{\frac{24}{25}} \]