ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios : $s i n 45^{\circ}$ is $\frac{1}{\sqrt{m}}$ , m is

Open in App

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, $S i n 45^{\circ} = S i n C = \frac{A B}{A C} = \frac{x}{x \sqrt{2}} = \frac{1}{\sqrt{2}}$

$∴ m = 2$

Suggest Corrections

Similar questions

c o s 45^{\circ}

\frac{m}{\sqrt{2}}

t a n 45^{\circ}

A B C

A B = B C = x

sin 45^{0}

\frac{1}{\sqrt{2}}

1

0

In the given figure, ABC is an isosceles right angle triangle, right angled at B. The ratio of the sides AB: BC : AC is _________.

Join BYJU'S Learning Program