Question

$ABC$ is an isosceles right-angled triangle. Assuming $AB=BC=x$, find the value of each of the following trigonometric ratios:

$\sin {45}^0$ is $\frac{1}{\sqrt{m}}$, the value of $m$ is

• A. 1
• B. 3
• C. 2
• D. 4

Answer 1

Answer: (C)

2

Given: $△ABC$, $AB=BC=x$, $\angle ABC={90}^{\circ }$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=x^2+x^2$
$A{C}^2=2x^2$
$AC=\sqrt{2}x$
Since, $AB=BC$, then by Isosceles triangle property,
$\angle ACB=\angle CAB={45}^{\circ }$
Now, $Sin{45}^{\circ }=SinC=\frac{P}{H}$

$=\frac{AB}{AC}$

$=\frac{x}{x\sqrt{2}}=\frac{1}{\sqrt{2}}$

so, the value of $m$ is $=2$