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Triangle ABC is an isosceles right triangle, right angled at B. Find the value of $\sin^2 A + \cos^2 C$.
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Solution: (a) $\triangle ABC$ is an isosceles right triangle
$\therefore \angle A = \angle C = 45^\circ$ (1/2 Mark)
$\sin^2 A + \cos^2 C = \sin^2 45^\circ + \cos^2 45^\circ$
$= \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2$ (1/2+1/2 Mark)
$= 1$ (1/2 Mark)
$\therefore \angle A = \angle C = 45^\circ$ (1/2 Mark)
$\sin^2 A + \cos^2 C = \sin^2 45^\circ + \cos^2 45^\circ$
$= \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2$ (1/2+1/2 Mark)
$= 1$ (1/2 Mark)