Consider the following for the three (03) items that follow: Let p...

Given p = \sin 35^{}^{\circ}, q = \sin 25^{}^{\circ}, and r = \sin(-95^{}^{\circ}) = -\sin(90^{}^{\circ} + 5^{}^{\circ}) = -\cos 5^{}^{\circ}.

p + q + r = \sin 35^{}^{\circ} + \sin 25^{}^{\circ} - \cos 5^{}^{\circ}

Using $\sin A + \sin B = 2 \sin$\left(\dfrac{A+B}{2}\right) \cos$\left(\dfrac{A-B}{2}\right):<br><br>$p + q + r = 2 \sin 30^{}^{\circ} \cos 5^{}^{\circ} - \cos 5^{}^{\circ} = 2 \left(\dfrac{1}{2}\right) \cos5^{}^{\circ} - \cos 5^{}^{\circ} = 0

We know that $(p + q + r)^2 = p² + q² + r² + 2(pq + qr + rp) = 0$

$pq + qr + rp =$-\dfrac{1}{2}($p^2 + q² + r²)$

Now, p^2 + q² + r² = \sin^2 35^{}^{\circ} + \sin^2 25^{}^{\circ} + \cos^2 5^{}^{\circ}

Using $\sin^2$\theta = \dfrac{1 - \cos 2\theta}{2}$and \cos^2$\theta = \dfrac{1 + \cos 2\theta}{2}:

$p^2 + q² + r² =$\dfrac{1 - \cos 70^{}^{\circ}}{2} + \dfrac{1 - \cos 50^{}^{\circ}}{2} + \dfrac{1 + \cos 10^{}^{\circ}}{2}

=\dfrac{3}{2} - \dfrac{1}{2}(\cos70^{}^{\circ} + \cos 50^{}^{\circ} - \cos 10^{}^{\circ})

Using $\cos C + \cos D = 2 \cos \left(\dfrac{C+D}{2}\right) \cos \left(\dfrac{C-D}{2}\right):<br><br>$\cos 70^{}^{\circ} + \cos 50^{}^{\circ} = 2 \cos 60^{}^{\circ} \cos 10^{}^{\circ} = 2 $\left(\dfrac{1}{2}\right) \cos$10^{}^{\circ} = \cos 10^{}^{\circ}$<br><br>So,$p^2 + q² + r² = $\dfrac{3}{2} - \dfrac{1}{2}(\cos$10^{}^{\circ} - \cos 10^{}^{\circ}) = $\dfrac{3}{2}<br><br>Therefore,$pq + qr + rp = $-\dfrac{1}{2} \left(\dfrac{3}{2}\right) = -\dfrac{3}{4}<br><br>Answer: -\dfrac{3}{4}$