Question

find the sine, cosine, and tangent of \\( \angle a \\).

simplify your answers and write them as proper fractions, improper fractions, or whole
numbers.

\\( \sin (a)= \\)
\\( \cos (a)= \\)
\\( \tan (a)= \\)

Explanation:

Step1: Find the length of side \(AB\)

Use the Pythagorean theorem \(AB=\sqrt{AC^{2}-BC^{2}}\). Given \(AC = 89\) and \(BC=80\), then \(AB=\sqrt{89^{2}-80^{2}}=\sqrt{(89 + 80)(89 - 80)}=\sqrt{169\times9}=\sqrt{1521}=39\).

Step2: Calculate \(\sin(A)\)

By the definition of sine in a right - triangle \(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite \(\angle A\) is \(BC = 80\), and the hypotenuse \(AC = 89\). So \(\sin(A)=\frac{80}{89}\).

Step3: Calculate \(\cos(A)\)

By the definition of cosine in a right - triangle \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The side adjacent to \(\angle A\) is \(AB = 39\), and the hypotenuse \(AC = 89\). So \(\cos(A)=\frac{39}{89}\).

Step4: Calculate \(\tan(A)\)

By the definition of tangent in a right - triangle \(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\). The side opposite \(\angle A\) is \(BC = 80\), and the side adjacent to \(\angle A\) is \(AB = 39\). So \(\tan(A)=\frac{80}{39}\).

Answer:

\(\sin(A)=\frac{80}{89}\), \(\cos(A)=\frac{39}{89}\), \(\tan(A)=\frac{80}{39}\)