Question

proving the law of sines two right triangles, △abd and △acd, were created from △abc by constructing ad ⊥ cb. follow these steps to prove that \\(\frac{\sin(b)}{b} = \frac{\sin(c)}{c}\\). 1. for △abd, \\(\sin(b) =\\) dropdown with options, image of triangle with labels a, b, c, d, h, b, c

Explanation:

Step1: Recall sine definition in right triangle

In a right triangle, the sine of an acute angle is the ratio of the length of the opposite side to the hypotenuse. For right triangle \( \triangle ABD \), angle \( B \) is an acute angle, the side opposite to angle \( B \) is \( AD \) (length \( h \)) and the hypotenuse is \( AB \) (length \( c \)). So, by the definition of sine, \( \sin(B)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{c} \)? Wait, no, wait. Wait, in \( \triangle ABD \), \( AD \perp CB \), so \( \triangle ABD \) is right-angled at \( D \). So angle at \( D \) is right angle. So angle \( B \): opposite side is \( AD = h \), hypotenuse is \( AB = c \)? Wait, no, wait the labels: in the diagram, \( AB \) is length \( c \), \( AC \) is length \( b \). Wait, maybe I mixed up. Wait, \( \triangle ABD \): right-angled at \( D \), so sides: \( AD = h \) (height), \( BD \) is adjacent to angle \( B \), \( AB = c \) (hypotenuse). So \( \sin(B)=\frac{AD}{AB}=\frac{h}{c} \)? Wait, but the options given are \( \frac{c}{h} \), \( \frac{h}{c} \), \( \frac{h}{b} \)? Wait, no, the user's image shows options: maybe the options are \( \frac{h}{c} \), \( \frac{c}{h} \), \( \frac{h}{b} \)? Wait, the problem says "For \( \triangle ABD \), \( \sin(B) = \) " with options (probably \( \frac{h}{c} \), \( \frac{c}{h} \), \( \frac{h}{b} \) – but looking at the diagram, \( AB = c \), \( AD = h \), right triangle at \( D \). So sine of angle \( B \) is opposite over hypotenuse: opposite side to \( B \) is \( AD = h \), hypotenuse is \( AB = c \). So \( \sin(B)=\frac{h}{c} \). Wait, but maybe I misread the labels. Wait, \( AC = b \), \( AB = c \), \( BC \) is the base, \( AD = h \) perpendicular to \( BC \). So in \( \triangle ABD \) (right-angled at \( D \)): angle at \( B \), so opposite side is \( AD = h \), hypotenuse is \( AB = c \). So \( \sin(B)=\frac{h}{c} \). Wait, but the options in the image: the blue box has options, maybe \( \frac{h}{c} \) is one of them. Wait, the user's problem step 1: "For \( \triangle ABD \), \( \sin(B) = \)" with options (probably \( \frac{h}{c} \), \( \frac{c}{h} \), \( \frac{h}{b} \)). So using the definition of sine in right triangle: \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). In \( \triangle ABD \), right-angled at \( D \), angle \( B \): opposite side is \( AD = h \), hypotenuse is \( AB = c \). So \( \sin(B)=\frac{h}{c} \).

Step2: Confirm the ratio

So applying the sine formula to \( \triangle ABD \), \( \sin(B) = \frac{h}{c} \).

Answer:

\(\frac{h}{c}\)