Question

1. a right triangle with a 45° angle, hypotenuse labeled 7, right angle, vertical leg labeled ( m ), horizontal leg labeled ( n ) (diagram included).

Explanation:

Step1: Identify the triangle type

This is a right - angled isosceles triangle (since one angle is \(45^{\circ}\), the other non - right angle is also \(45^{\circ}\) as the sum of angles in a triangle is \(180^{\circ}\), so \(180 - 90 - 45=45^{\circ}\)). In a right - angled isosceles triangle, the legs are equal (\(m = n\)) and we can use trigonometric ratios or the Pythagorean theorem. We can also use the sine or cosine function. Let's use the sine function for side \(m\). The sine of an angle in a right triangle is defined as \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\theta = 45^{\circ}\), the opposite side to \(45^{\circ}\) is \(m\) and the hypotenuse is \(7\). So \(\sin(45^{\circ})=\frac{m}{7}\).
We know that \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\), so \(m = 7\times\sin(45^{\circ})=7\times\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}\approx\frac{7\times1.414}{2}=\frac{9.898}{2} = 4.949\). Similarly, for side \(n\), using \(\cos(45^{\circ})=\frac{n}{7}\), and since \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), we get \(n=\frac{7\sqrt{2}}{2}\) as well.

Step2: (Optional, if we want to use Pythagorean theorem)

In a right - angled triangle, \(m^{2}+n^{2}=7^{2}\). But since \(m = n\) (isosceles right triangle), we have \(2m^{2}=49\), so \(m^{2}=\frac{49}{2}\), and \(m=\sqrt{\frac{49}{2}}=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\) (rationalizing the denominator), same as before.

Answer:

If we are finding the length of \(m\) (or \(n\)): \(\frac{7\sqrt{2}}{2}\) (or approximately \(4.95\))