Question

solve for x

1. solve for x
2. solve for x

solve for x

1. solve for x
2. solve for x

2sqrt(15)
solve for x

1. solve for x
2. solve for x

solve for x

1. solve for x
2. solve for x

Explanation:

Step1: Identify the triangle type (First triangle: 30 - 60 - 90)

In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\). The side opposite 30° is the shortest, then opposite 60° (longer leg), then hypotenuse (twice the shortest). Here, the shorter leg (opposite 30°) is 8? Wait, no, the first triangle: right angle, 30°, so angle 30°, right angle, so the side opposite 30° is \(x\)? Wait, no, the given side is 8 (adjacent to 30°? Wait, no, let's re - examine. The first triangle: right angle, angle 30°, side adjacent to 30° is 8? Wait, no, the side labeled 8 is the longer leg (opposite 60°), and \(x\) is the shorter leg (opposite 30°). In 30 - 60 - 90, longer leg \(=\) shorter leg \(\times\sqrt{3}\). So if longer leg is 8, then shorter leg \(x=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\)? Wait, maybe I misread. Wait, the first triangle: angle 30°, right angle, so the side opposite 30° is \(x\), and the side adjacent (longer leg) is 8. So \( \tan(30^{\circ})=\frac{x}{8}\), \(x = 8\times\tan(30^{\circ})=8\times\frac{1}{\sqrt{3}}=\frac{8\sqrt{3}}{3}\). But maybe let's take another triangle, like problem 3: isosceles right triangle (legs equal, angle 45°), hypotenuse \(2\sqrt{2}\). In isosceles right triangle, legs \(x\), hypotenuse \(x\sqrt{2}\). So \(x\sqrt{2}=2\sqrt{2}\), so \(x = 2\).

Step2: Solve for \(x\) in problem 3 (isosceles right triangle)

For an isosceles right triangle (45 - 45 - 90), the hypotenuse \(h=x\sqrt{2}\), where \(x\) is the length of each leg. Given hypotenuse \(h = 2\sqrt{2}\).
We have the formula \(h=x\sqrt{2}\). Substitute \(h = 2\sqrt{2}\) into the formula:

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Answer:

For problem 3, \(x = 2\)