Question

find the sine, cosine, and tangent of ∠i. write your answer in simplified, rationalized form. do not round. sin (i) = cos (i) = tan (i) =

Explanation:

Step1: Find the hypotenuse

Use the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 4\) and \(b=9\). So \(c=\sqrt{4^{2}+9^{2}}=\sqrt{16 + 81}=\sqrt{97}\).

Step2: Calculate sine

By the definition of sine in a right - triangle \(\sin(I)=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(\angle I\) is \(GH = 4\) and the hypotenuse is \(\sqrt{97}\), so \(\sin(I)=\frac{4}{\sqrt{97}}=\frac{4\sqrt{97}}{97}\).

Step3: Calculate cosine

By the definition of cosine in a right - triangle \(\cos(I)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(\angle I\) is \(GI = 9\) and the hypotenuse is \(\sqrt{97}\), so \(\cos(I)=\frac{9}{\sqrt{97}}=\frac{9\sqrt{97}}{97}\).

Step4: Calculate tangent

By the definition of tangent in a right - triangle \(\tan(I)=\frac{\text{opposite}}{\text{adjacent}}\). The opposite side to \(\angle I\) is \(4\) and the adjacent side is \(9\), so \(\tan(I)=\frac{4}{9}\).

Answer:

\(\sin(I)=\frac{4\sqrt{97}}{97}\), \(\cos(I)=\frac{9\sqrt{97}}{97}\), \(\tan(I)=\frac{4}{9}\)