Question

the law of sines
two right triangles, △abd and △acd, were created from △abc by constructing $overline{ad} \perp \overline{cb}$.
follow these steps to prove that $\frac{\sin(b)}{b} = \frac{\sin(c)}{c}$.

1. multiply both sides of the equation by $c$ and simplify. the equation $\sin(b) = \frac{h}{c}$ becomes $c \sin(b) = hw$.
2. for △acd, $\sin(c) = h/w$
3. multiply both sides of the equation by $b$ and simplify. the equation $\sin(c) = \frac{h}{b}$ becomes dropdown with options $b \sin(c) = h$, $h \sin(c) = h$

check

Explanation:

Step 2 Explanation:

We start with the equation $\sin(B) = \frac{h}{c}$. To eliminate the denominator $c$, we multiply both sides of the equation by $c$.

Multiplying the left side: $c \times \sin(B) = c\sin(B)$

Multiplying the right side: $c \times \frac{h}{c} = h$ (the $c$ in the numerator and denominator cancels out)

So the equation becomes $c\sin(B) = h$.

Step 3 Explanation:

In right triangle $\triangle ACD$, $\angle ADC = 90^\circ$ (since $AD \perp CB$). The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

For $\angle C$ in $\triangle ACD$:

• The opposite side to $\angle C$ is $AD$, which has length $h$.
• The hypotenuse is $AC$, which has length $b$ (from the diagram, $AC = b$).

So by the definition of sine, $\sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{b}$.

Step 4 Explanation:

We start with the equation $\sin(C) = \frac{h}{b}$. To eliminate the denominator $b$, we multiply both sides of the equation by $b$.

Multiplying the left side: $b \times \sin(C) = b\sin(C)$

Multiplying the right side: $b \times \frac{h}{b} = h$ (the $b$ in the numerator and denominator cancels out)

So the equation becomes $b\sin(C) = h$.

Final Answer for Step 2:

$c\sin(B) = h$

Final Answer for Step 3:

$\frac{h}{b}$

Final Answer for Step 4:

$b\sin(C) = h$

Answer:

Step 2 Explanation:

We start with the equation $\sin(B) = \frac{h}{c}$. To eliminate the denominator $c$, we multiply both sides of the equation by $c$.

Multiplying the left side: $c \times \sin(B) = c\sin(B)$

Multiplying the right side: $c \times \frac{h}{c} = h$ (the $c$ in the numerator and denominator cancels out)

So the equation becomes $c\sin(B) = h$.

Step 3 Explanation:

In right triangle $\triangle ACD$, $\angle ADC = 90^\circ$ (since $AD \perp CB$). The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

For $\angle C$ in $\triangle ACD$:

• The opposite side to $\angle C$ is $AD$, which has length $h$.
• The hypotenuse is $AC$, which has length $b$ (from the diagram, $AC = b$).

So by the definition of sine, $\sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{b}$.

Step 4 Explanation:

We start with the equation $\sin(C) = \frac{h}{b}$. To eliminate the denominator $b$, we multiply both sides of the equation by $b$.

Multiplying the left side: $b \times \sin(C) = b\sin(C)$

Multiplying the right side: $b \times \frac{h}{b} = h$ (the $b$ in the numerator and denominator cancels out)

So the equation becomes $b\sin(C) = h$.

Final Answer for Step 2:

$c\sin(B) = h$

Final Answer for Step 3:

$\frac{h}{b}$

Final Answer for Step 4:

$b\sin(C) = h$