It is well known that $\sin 45=\frac{\sqrt{2}}{2}$ . So $x=45°$ . The triangle is isosceles, meaning that angle $ACB$ is $45°$ too, hence angle $BAC=90°$ .

We now know that this is a right-angled triangle.

Let the side length $AC$ be known as $y$ , for now. Using Pythagoras:

$y^2+y^2=(\sqrt{2})^2$

$2y^2=2$

$y^2=1$

$y=1$

So the answer to this question is 1.

• FInding angle $x$

$\sin{(x)} = \dfrac{\sqrt{2}}{2} \implies x = 45^{\circ} \quad [\because x < 180^{\circ}]$

• Since it's an isosceles triangle

$\angle ACB = 45^{\circ} \implies \angle BAC = 90^{\circ}$

• Now using Pythagoras' theorem

$\begin{aligned} AC^2 + AB^2 &= (\sqrt{2})^2 \\ 2 AC^2 &= 2 \\ AC &= \boxed{1} \end{aligned}$

$\sin45^{\circ}=\cfrac{\sqrt{2}}{2}$ This is a right triangle, so $2a^2=\sqrt{2}^2$ , so $a=1$ .

$sin x$ = $sqrt{2}$ /2

So cos x= sqrt{(1-(\(sin^2 (x) ))}= $sqrt{(1-(2/4))}$ = $sqrt(2/4)$ = $sqrt{2}$ /2

Let AC=AB=y Now by Law of Cosine (AC)^2=((AB)^2)+((BC)^2)-(2xABxBCx(cos(x)))

So (y^2) = (y^2) + 2 -(2 sqrt(2) y*(sqrt(2)/2)) 2y=2

Giving y=1 So AC=1

Here is a method of solving the problem without finding the angle x.

First, let's add a midpoint $M$ of Segment $BC$ . With $AM$ perpendicular to $BC$ , you get two identical right triangles $\bigtriangleup ABM$ and $\bigtriangleup ACM$ .

Then, we find the measure of $BM$

$BM=CM$

$BC=BM+CM$

$BC=2BM$

$BM=\frac{\sqrt{2}}{2}$ .

With that, by definition of sine,

$\sin(x)=\frac{BM}{AB}$

$\frac{\sqrt{2}}{2}=\frac{\sqrt{2}/2}{AB}$

$AB=\frac{\sqrt{2}/2}{\sqrt{2}/2}$

And finally, $AB=AC=1$

\sinx = \frac{opposite catheter}{hypotenuse} = \frac {\overline{AC}}{\sqrt{2}} = \frac {\sqrt{2}}{2} _therefore_ \overline{AC} = \frac {\sqrt{2} * \sqrt{2}}{2} = \frac {2}{2} =1

We need not consider anything but the fact that triangle is isosceles, let the missing length $AC$ be $a$ , by Pythagoras theorem,

$a²+a²= (\sqrt{2})^2 \rightarrow a²=1$

Or $a=1$

$\sin x = \dfrac{\sqrt{2}}{2}$ $\Rightarrow$ $x=45^\circ$

By law of sines,

$\dfrac{\sin \angle ABC}{AC}=\dfrac{\sin \angle BAC}{BC}$

$\dfrac{\sin x}{AC}=\dfrac{\sin (180-2x)}{\sqrt{2}}$

$\dfrac{\sin 45}{AC}=\dfrac{\sin 90}{\sqrt{2}}$

$AC=\dfrac{\sqrt{2}\times \sin 45}{\sin 90}=\color{#3D99F6}\boxed{1}$ $\color{#EC7300}\boxed{answer}$

$\sin x = \frac{1}{√2}$ So, we know if $\sin \theta=\frac{1}{\sqrt{2}}$ then , $\theta =45°\pm 2\pi n$

Now as in this case, $0°\le x\leq 90°$

So, $x=45°$ So, $\angle BAC= 90°$

Now by Sine Rule,we know $\frac{AC}{\sin 45°}=\frac{BC}{\sin 90°}$

$\rightarrow \frac{AC}{\frac{1}{\sqrt{2}}}=\sqrt {2}$

$\rightarrow AC= 1 units$

In a right isosceles triangle, if the legs are x, then the hypotenuse is x $\sqrt { 2 }$ . So in this question, we are given the hypotenuse, so the legs are $\frac{\sqrt { 2 }}{\sqrt { 2 }}$ , which is 1.

who else had come from his link n todays (26 or 25/7/20)???