Question

determine \\(\sin(a)\\) and \\(\sin(b)\\). image of a right triangle with right angle at c, side ac = 21, side bc = 20, side ab = 29

Explanation:

Step1: Recall Sine Definition

In a right triangle, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. Triangle \(ABC\) is right - angled at \(C\), so hypotenuse \(AB\) can be found using Pythagorean theorem? Wait, no, we can see the sides: \(AC = 21\), \(BC = 20\), \(AB\)? Wait, no, in right - triangle \(ABC\) with right angle at \(C\), for angle \(A\), the opposite side is \(BC\) and hypotenuse is \(AB\). For angle \(B\), the opposite side is \(AC\) and hypotenuse is \(AB\). First, find hypotenuse \(AB\)? Wait, no, wait the sides: \(AC = 21\), \(BC = 20\), and \(AB\) is the hypotenuse. Wait, but we can check if \(21 - 20 - 29\) is a Pythagorean triple: \(21^{2}+20^{2}=441 + 400=841\), and \(29^{2}=841\). So hypotenuse \(AB = 29\).

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

For angle \(A\), opposite side to \(A\) is \(BC = 20\), hypotenuse \(AB = 29\). So \(\sin(A)=\frac{\text{opposite to }A}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{20}{29}\).

Step3: Calculate \(\sin(B)\)

For angle \(B\), opposite side to \(B\) is \(AC = 21\), hypotenuse \(AB = 29\). So \(\sin(B)=\frac{\text{opposite to }B}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{21}{29}\).

Answer:

\(\sin(A)=\frac{20}{29}\), \(\sin(B)=\frac{21}{29}\)