Question

solve for x
(right triangle with hypotenuse 20, one angle 45°, right angle, and base x)

Explanation:

Step1: Identify triangle type

The triangle is right - angled with one angle \(45^{\circ}\), so it's a 45 - 45 - 90 triangle (isosceles right triangle), or we can use trigonometry. Let's use cosine: \(\cos(45^{\circ})=\frac{x}{20}\).

Step2: Recall \(\cos(45^{\circ})\) value

We know that \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071\).

Step3: Solve for \(x\)

From \(\cos(45^{\circ})=\frac{x}{20}\), we get \(x = 20\times\cos(45^{\circ})\). Substituting \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), we have \(x = 20\times\frac{\sqrt{2}}{2}=10\sqrt{2}\approx14.14\). Also, since it's a 45 - 45 - 90 triangle, the legs are equal, and using Pythagorean theorem: \(x^{2}+x^{2}=20^{2}\), \(2x^{2} = 400\), \(x^{2}=200\), \(x=\sqrt{200}=10\sqrt{2}\approx14.14\).

Answer:

\(x = 10\sqrt{2}\) (or approximately \(14.14\))