Question

1. a right triangle with hypotenuse ( 2sqrt{6} ), one acute angle ( 45^circ ), legs labeled ( y ) and ( x ), right angle at the vertex between ( x ) and ( y ).

Explanation:

Step1: Identify the triangle type

This is a right - angled isosceles triangle? Wait, no, one angle is \(45^{\circ}\), right angle is \(90^{\circ}\), so the third angle is also \(45^{\circ}\)? Wait, no, in a right - angled triangle, if one non - right angle is \(45^{\circ}\), then the other non - right angle is \(180 - 90 - 45=45^{\circ}\), so it is an isosceles right - angled triangle? Wait, no, the hypotenuse is \(2\sqrt{6}\), and we can use trigonometric ratios. Let's recall that in a right - angled triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and also for a right - angled triangle with angles \(45^{\circ},45^{\circ},90^{\circ}\), the legs are equal, and hypotenuse \(h = l\sqrt{2}\), where \(l\) is the length of the leg.

Wait, let's check. Let's assume the right - angled triangle has legs \(x\) and \(y\), hypotenuse \(c = 2\sqrt{6}\), angle \(\theta = 45^{\circ}\) adjacent to side \(y\) and opposite to side \(x\).

Using the Pythagorean theorem: \(x^{2}+y^{2}=c^{2}=(2\sqrt{6})^{2}=24\)

Also, since one angle is \(45^{\circ}\), \(\tan(45^{\circ})=\frac{x}{y}=1\), so \(x = y\)

Substitute \(x = y\) into the Pythagorean theorem: \(x^{2}+x^{2}=24\)

Step2: Solve for \(x\) (and \(y\) since \(x = y\))

\(2x^{2}=24\)

Divide both sides by 2: \(x^{2}=12\)

Take the square root: \(x=\sqrt{12}=2\sqrt{3}\)

Since \(x = y\), \(y = 2\sqrt{3}\)

Alternatively, using trigonometric ratios:

\(\sin(45^{\circ})=\frac{x}{2\sqrt{6}}\)

We know that \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\), so \(\frac{\sqrt{2}}{2}=\frac{x}{2\sqrt{6}}\)

Cross - multiply: \(x=\frac{\sqrt{2}}{2}\times2\sqrt{6}=\sqrt{2}\times\sqrt{6}=\sqrt{12}=2\sqrt{3}\)

Similarly, \(\cos(45^{\circ})=\frac{y}{2\sqrt{6}}\), and \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), so \(y=\frac{\sqrt{2}}{2}\times2\sqrt{6}=2\sqrt{3}\)

Answer:

\(x = 2\sqrt{3}\), \(y = 2\sqrt{3}\)