Question

find the length of side $x$ to the nearest tenth.
(there is a right triangle with a hypotenuse of $sqrt{6}$ and two acute angles of $45^circ$ each, and one leg is $x$)

Explanation:

Step1: Identify trigonometric ratio

We use sine: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
$\sin(45^\circ)=\frac{x}{\sqrt{6}}$

Step2: Solve for $x$

Rearrange to isolate $x$: $x = \sqrt{6} \times \sin(45^\circ)$
Substitute $\sin(45^\circ)=\frac{\sqrt{2}}{2}$:
$x = \sqrt{6} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$

Step3: Calculate decimal value

$\sqrt{3} \approx 1.732$
Round to nearest tenth: $x \approx 1.7$

Answer:

1.7