Question

find the sine, cosine, and tangent of ∠k. simplify your answers and write them as proper fractions, improper fractions numbers. sin (k) = cos (k) = tan (k) =

Explanation:

Step1: Find the length of JK

In right triangle \(IJK\) with right angle at \(J\), we can use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 17\) (hypotenuse \(IK\)) and \(a = 15\) (leg \(IJ\)). Let \(JK=b\). Then:
\[
b^{2}+15^{2}=17^{2}
\]
\[
b^{2}=17^{2}- 15^{2}
\]
\[
b^{2}=289 - 225
\]
\[
b^{2}=64
\]
\[
b = 8
\]
So, \(JK = 8\), \(IJ=15\), \(IK = 17\).

Step2: Find \(\sin(K)\)

For an acute angle \(K\) in a right triangle, \(\sin(K)=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite \(\angle K\) is \(IJ = 15\), and the hypotenuse is \(IK=17\). So:
\[
\sin(K)=\frac{15}{17}
\]

Step3: Find \(\cos(K)\)

For an acute angle \(K\) in a right triangle, \(\cos(K)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The side adjacent to \(\angle K\) is \(JK = 8\), and the hypotenuse is \(IK = 17\). So:
\[
\cos(K)=\frac{8}{17}
\]

Step4: Find \(\tan(K)\)

For an acute angle \(K\) in a right triangle, \(\tan(K)=\frac{\text{opposite}}{\text{adjacent}}\). The side opposite \(\angle K\) is \(IJ = 15\), and the side adjacent to \(\angle K\) is \(JK = 8\). So:
\[
\tan(K)=\frac{15}{8}
\]

Answer:

\(\sin(K)=\frac{15}{17}\), \(\cos(K)=\frac{8}{17}\), \(\tan(K)=\frac{15}{8}\)