ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios : $c o s 45^{\circ}$ is $\frac{m}{\sqrt{2}}$ , m is

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Solution

Given: $△ A B C$ , $A B = B C = x$ , $∠ A B C = 90^{\circ}$
By Pythagoras Theorem,
$A C^{2} = A B^{2} + B C^{2}$
$A C^{2} = x^{2} + x^{2}$
$A C^{2} = 2 x^{2}$
$A C = \sqrt{2} x$
Since, $A B = B C$ , then by Isosceles triangle property,
$∠ A C B = ∠ C A B = 45^{\circ}$
Now, $c o s 45^{\circ} = c o s C = \frac{B}{H} = \frac{B C}{A C} = \frac{x}{x \sqrt{2}} = \frac{1}{\sqrt{2}}$

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