Question

which one is the correct way to solve x
sin(90)=x/8
cos(90) = x/8
x=√(6²+8²)

Explanation:

Step1: Analyze the triangle

The triangle is a right - triangle with legs of length 6 and 8, and hypotenuse \(x\). We can use the Pythagorean theorem which states that for a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(c = \sqrt{a^{2}+b^{2}}\). Also, recall the values of \(\sin(90^{\circ})\) and \(\cos(90^{\circ})\): \(\sin(90^{\circ})=1\) and \(\cos(90^{\circ}) = 0\).

Step2: Analyze Option 1 (\(\sin(90)=\frac{x}{8}\))

If we substitute \(\sin(90^{\circ}) = 1\) into the equation \(\sin(90^{\circ})=\frac{x}{8}\), we get \(1=\frac{x}{8}\), so \(x = 8\). But if we use the Pythagorean theorem, \(x=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}=10\), so this option is incorrect.

Step3: Analyze Option 2 (\(\cos(90)=\frac{x}{8}\))

Since \(\cos(90^{\circ})=0\), substituting into the equation \(\cos(90^{\circ})=\frac{x}{8}\) gives \(0=\frac{x}{8}\), so \(x = 0\), which is impossible for the hypotenuse of a right - triangle with legs 6 and 8, so this option is incorrect.

Step4: Analyze Option 3 (\(x=\sqrt{6^{2}+8^{2}}\))

For a right - triangle, the Pythagorean theorem is \(c^{2}=a^{2}+b^{2}\), where \(c\) is the hypotenuse and \(a\) and \(b\) are the legs. Here, \(a = 6\), \(b = 8\) and \(c=x\). So \(x^{2}=6^{2}+8^{2}\), and taking the square root of both sides, we get \(x=\sqrt{6^{2}+8^{2}}\). Calculating \(6^{2}=36\), \(8^{2}=64\), then \(6^{2}+8^{2}=36 + 64=100\), and \(\sqrt{100}=10\), which is the correct length of the hypotenuse.

Answer:

The correct way to solve for \(x\) is \(x=\sqrt{6^{2}+8^{2}}\) (the third option, the green - colored option with the formula \(x=\sqrt{6^{2}+8^{2}}\)).