Question

question 12 of 20
consider the right triangle given below. the length of side b to two decimal
places is ____.
image of right triangle with right angle at c, hypotenuse ab = 25, angle at b is 65 degrees, side ac is b, side bc is a
a. 10.57
b. 22.66
c. 14.06
d. 20.67

Explanation:

Step1: Identify the trigonometric relation

In right triangle \(ABC\) (right-angled at \(C\)), the hypotenuse \(AB = 25\), angle at \(B\) is \(65^\circ\). We need to find side \(b\) (which is opposite to angle \(B\)? Wait, no, side \(b\) is \(AC\), adjacent to angle \(B\)? Wait, no, let's clarify: in right triangle, for angle \(B\), the sides: hypotenuse \(AB = 25\), side \(AC = b\) (opposite to angle \(B\))? Wait, no, angle at \(B\) is \(65^\circ\), so angle \(B\): the sides: adjacent side is \(BC = a\), opposite side is \(AC = b\), hypotenuse \(AB = 25\). So \(\sin(65^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{25}\), so \(b = 25\times\sin(65^\circ)\). Wait, or \(\cos(65^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\), if \(b\) is adjacent? Wait, no, let's draw: points \(A\), \(B\), \(C\) with right angle at \(C\). So \(AC\) is horizontal (side \(b\)), \(BC\) is vertical (side \(a\)), \(AB\) is hypotenuse (25). Angle at \(B\) is \(65^\circ\), so angle between \(AB\) and \(BC\) is \(65^\circ\). So in triangle, angle at \(B\): the side opposite to angle \(B\) is \(AC = b\), adjacent is \(BC = a\), hypotenuse \(AB = 25\). So \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(\sin(65^\circ)=\frac{b}{25}\), so \(b = 25\times\sin(65^\circ)\).

Step2: Calculate \(\sin(65^\circ)\)

We know that \(\sin(65^\circ)\approx0.9063\) (using calculator, make sure it's in degree mode).

Step3: Compute \(b\)

\(b = 25\times0.9063 = 22.6575\), which rounds to \(22.66\) when rounded to two decimal places.

Answer:

B. 22.66