Question

select the correct answer from each drop - down menu. consider right triangle abc. posttest: right triangles and trigonometry 1 1 sin(a)= cos(a)=

Explanation:

Step1: Recall the definitions of sine and cosine in a right - triangle

In a right - triangle, for an acute angle \(A\), \(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\).
For angle \(A\) in right - triangle \(ABC\), the side opposite to angle \(A\) is \(BC = 9\), the side adjacent to angle \(A\) is \(AB=40\), and the hypotenuse \(AC = 41\).

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

Using the formula \(\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\), we substitute the values. So, \(\sin(A)=\frac{BC}{AC}\).
Since \(BC = 9\) and \(AC = 41\), \(\sin(A)=\frac{9}{41}\).

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

Using the formula \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\), we substitute the values. So, \(\cos(A)=\frac{AB}{AC}\).
Since \(AB = 40\) and \(AC = 41\), \(\cos(A)=\frac{40}{41}\).

Answer:

\(\sin(A)=\frac{BC}{AC}=\frac{9}{41}\), \(\cos(A)=\frac{AB}{AC}=\frac{40}{41}\)