Question

find sin(\beta) in the triangle. choose 1 answer: \circled{1} \frac{12}{13} \circled{2} \frac{12}{5} \circled{3} \frac{5}{12} \circled{4} \frac{5}{13} (diagram: triangle with right angle at c, vertices b, c, a; sides bc=5, ac=12, ab=13)

Explanation:

Step1: Recall sine definition

In a right triangle, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$.

Step2: Identify sides for $\beta$

For angle $\beta$ (at vertex $B$), the opposite side is $AC = 12$, and the hypotenuse is $AB = 13$? Wait, no, wait. Wait, triangle $BCA$: right angle at $C$. Wait, angle at $B$: $\beta$. So in triangle $ABC$, right - angled at $C$? Wait, no, the right angles: $\angle C$ and $\angle B$? Wait, no, the diagram: $\angle C$ is right angle, $\angle B$ is another right angle? Wait, no, maybe it's a quadrilateral? Wait, no, looking at the triangle: sides $BC = 5$, $AC = 12$, $AB = 13$? Wait, no, the labels: $BC = 5$, $AC = 12$, $AB = 13$? Wait, angle $\beta$ is at $B$. So in triangle $ABC$, with right angle at $C$? Wait, no, the right angles: $\angle C$ and $\angle B$ are right angles? That can't be. Wait, maybe it's a right triangle at $C$ and another right angle at $B$? No, that's impossible. Wait, maybe the triangle is right - angled at $C$, and angle $\beta$ is at $B$. So sides: $BC = 5$ (adjacent to $\beta$), $AC = 12$ (opposite to $\beta$), and $AB$ is the hypotenuse. Wait, by Pythagoras, $AB=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$. So for angle $\beta$ (at $B$), in right - triangle (right - angled at $C$), $\sin(\beta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}$? Wait, no, wait: opposite side to $\beta$: when angle is at $B$, the opposite side is $AC$, hypotenuse is $AB$. Wait, but let's re - check. If $\angle C$ is right angle, then $AB$ is hypotenuse, $BC = 5$, $AC = 12$, $AB = 13$. Angle at $B$: $\beta$. So $\sin(\beta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}$? Wait, no, wait: in angle $\beta$, the sides: adjacent is $BC$, opposite is $AC$, hypotenuse is $AB$. So $\sin(\beta)=\frac{AC}{AB}=\frac{12}{13}$. Wait, but let's check the options. Option a is $\frac{12}{13}$, option d is $\frac{5}{13}$. Wait, maybe I made a mistake. Wait, maybe the right angle is at $B$? Wait, the diagram shows right angles at $C$ and $B$? No, that's a rectangle. Wait, no, it's a triangle. Wait, maybe the triangle is right - angled at $B$, so $\angle B = 90^{\circ}$. Then, sides: $BC = 5$, $AB = 13$, $AC = 12$ (by Pythagoras: $5^{2}+12^{2}=25 + 144 = 169=13^{2}$). So angle $\beta$ is at $B$? No, if $\angle B$ is right angle, then angle $\beta$ is another angle. Wait, maybe the angle $\beta$ is at $B$, and the triangle is right - angled at $C$. So angle at $B$: $\beta$, right angle at $C$. So sides: $BC = 5$ (leg), $AC = 12$ (leg), $AB = 13$ (hypotenuse). Then, $\sin(\beta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}$? Wait, but let's think again. In angle $\beta$ (at $B$), the opposite side is $AC$, hypotenuse is $AB$. So $\sin(\beta)=\frac{AC}{AB}=\frac{12}{13}$. So the correct option is the first one? Wait, no, wait: maybe I mixed up opposite and adjacent. Wait, if angle is at $B$, then the sides: $BC$ is adjacent to $\beta$, $AC$ is opposite, and $AB$ is hypotenuse. So $\sin(\beta)=\frac{opposite}{hypotenuse}=\frac{AC}{AB}=\frac{12}{13}$. So the answer should be $\frac{12}{13}$.

Answer:

$\frac{12}{13}$ (corresponding to the first option, e.g., if the first option is A. $\frac{12}{13}$, then the answer is A. $\frac{12}{13}$)