Question

which one is the correct way to solve x
$x = \sqrt{6^2 + 8^2}$
$\cos(90) = x/8$
$\sin(90)=x/8$

Explanation:

Step1: Identify the triangle type

The triangle is a right - triangle with legs of length 6 and 8, and hypotenuse \(x\).

Step2: Recall the Pythagorean theorem

For a right - triangle, the Pythagorean theorem states that \(a^{2}+b^{2}=c^{2}\), where \(a\) and \(b\) are the lengths of the legs and \(c\) is the length of the hypotenuse. Here, \(a = 6\), \(b = 8\) and \(c=x\). So, \(x=\sqrt{6^{2}+8^{2}}\).

Step3: Analyze the trigonometric options

• For the blue option: \(\cos(90^{\circ}) = 0\), and if \(\cos(90^{\circ})=\frac{x}{8}\), then \(0=\frac{x}{8}\) which implies \(x = 0\), which is not possible for the hypotenuse of a right - triangle with legs 6 and 8.
• For the green option: \(\sin(90^{\circ})=1\), and if \(\sin(90^{\circ})=\frac{x}{8}\), then \(1=\frac{x}{8}\) which implies \(x = 8\), but the hypotenuse of a right - triangle with legs 6 and 8 should be \(\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100} = 10

eq8\).

Answer:

\(x=\sqrt{6^{2}+8^{2}}\) (the orange - colored option)