Question

find the length of side x to the nearest tenth. triangle with right angle, two 45° angles, hypotenuse √7, side x opposite one 45° angle

Explanation:

Step1: Identify trigonometric ratio

We use sine: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
Here, $\theta=45^\circ$, opposite side is $x$, hypotenuse is $\sqrt{7}$.

Step2: Substitute values and solve

$$\sin(45^\circ)=\frac{x}{\sqrt{7}}$$
We know $\sin(45^\circ)=\frac{\sqrt{2}}{2}$, so:
$$x=\sqrt{7} \times \frac{\sqrt{2}}{2}$$

Step3: Calculate and round

First compute $\sqrt{7}\approx2.6458$, $\sqrt{2}\approx1.4142$:
$$x\approx\frac{2.6458\times1.4142}{2}\approx\frac{3.7417}{2}\approx1.8708$$
Round to nearest tenth: $x\approx1.9$

Answer:

$1.9$