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if, in a right triangle, $\theta = 26^{circ}$ and the side adjacent to …

Question

if, in a right triangle, $\theta = 26^{circ}$ and the side adjacent to $\theta$ has length 6, how long is the hypotenuse? round to at least 3 decimal places.

Explanation:

Step1: Recall the cosine formula

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Let the hypotenuse be \(h\). Given \(\theta = 26^{\circ}\) and adjacent \(= 6\). So, \(\cos(26^{\circ})=\frac{6}{h}\).

Step2: Solve for \(h\)

Rearrange the formula to \(h=\frac{6}{\cos(26^{\circ})}\). We know that \(\cos(26^{\circ})\approx0.898794\). Then \(h=\frac{6}{0.898794}\).

Step3: Calculate the value of \(h\)

\(h=\frac{6}{0.898794}\approx6.675\)

Answer:

\(6.675\)