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In the image of triangle [tex]\( \triangle ABC \)[/tex], [tex]\( b \)[/tex] is the length between points [tex]\( A \)[/tex] and [tex]\( C \)[/tex], [tex]\( a \)[/tex] is the length between points [tex]\( C \)[/tex] and [tex]\( B \)[/tex], and [tex]\( c \)[/tex] is the base length between points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. [tex]\( D \)[/tex] is a point on [tex]\( AB \)[/tex] forming a right angle [tex]\( DCB \)[/tex].

Proof:
[tex]\[
\begin{array}{|l|l|}
\hline
\text{Statements} & \text{Reasons} \\
\hline
1. \overline{CD} \text{ is an altitude of } \triangle ABC. & \text{Given} \\
\hline
2. \angle ADC \text{ and } \angle BDC \text{ are right angles.} & \text{Definition of altitude} \\
\hline
3. \triangle ADC \text{ and } \triangle BCD \text{ are right triangles.} & \text{Definition of right triangles} \\
\hline
4. \sin(A) = \frac{CD}{b} \text{ and } \sin(B) = \frac{CD}{a} & \text{Definition of sine} \\
\hline
5. & \text{Multiplication property of equality} \\
\hline
6. b \sin(A) = a \sin(B) & \text{Substitution property of equality} \\
\hline
7. \frac{a}{\sin(A)} = \frac{b}{\sin(B)} & \text{Division property of equality} \\
\hline
\end{array}
\][/tex]

Which statement completes this proof?

A. [tex]\( CD = b \sin(A) \)[/tex] and [tex]\( CD = a \sin(B) \)[/tex]

B. [tex]\( CD = b \sin(B) \)[/tex] and [tex]\( CD = a \sin(A) \)[/tex]

C. [tex]\( b = CD \sin(A) \)[/tex] and [tex]\( a = CD \sin(B) \)[/tex]

D. [tex]\( b = CD \sin(B) \)[/tex] and [tex]\( a = CD \sin(A) \)[/tex]



Answer :

GinnyAnswer
To determine which statement completes this proof, let's analyze the given steps in detail:

1. [tex]$\overline{CD}$[/tex] is an altitude of [tex]$\triangle ABC$[/tex].
Reason: Given.

2. [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles.
Reason: Definition of altitude (an altitude creates a right angle with the base of the triangle).

3. [tex]$\triangle ADC$[/tex] and [tex]$\triangle BCD$[/tex] are right triangles.
Reason: Both angles [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles, thus forming right triangles.

4. [tex]$\sin(A) = \frac{CD}{b}$[/tex] and [tex]$\sin(B) = \frac{CD}{a}$[/tex].
Reason: Definition of sine in right triangles:
- [tex]$\sin(A)$[/tex] is the ratio of the length of the side opposite angle A ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle ADC$[/tex] ([tex]$b$[/tex]).
- Similarly, [tex]$\sin(B)$[/tex] is the ratio of the length of the side opposite angle B ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle BCD$[/tex] ([tex]$a$[/tex]).

5. Multiplication property of equality:
From the equations in step 4, we can derive:
- Multiplying both sides of [tex]$\sin(A) = \frac{CD}{b}$[/tex] by [tex]$b$[/tex], we get [tex]$b \cdot \sin(A) = CD$[/tex].
- Multiplying both sides of [tex]$\sin(B) = \frac{CD}{a}$[/tex] by [tex]$a$[/tex], we get [tex]$a \cdot \sin(B) = CD$[/tex].

6. [tex]$CD = b \cdot \sin(A)$[/tex] and [tex]$CD = a \cdot \sin(B)$[/tex].
Reason: From the multiplication property of equality, we get these two relations.

7. By the Substitution property of equality, we equate the two expressions for [tex]$CD$[/tex]:
[tex]$b \cdot \sin(A) = a \cdot \sin(B)$[/tex].
- Dividing both sides by [tex]$\sin(A) \cdot \sin(B)$[/tex], we get [tex]$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$[/tex].

8. Finally, we conclude that:
- [tex]$\boxed{b = CD \cdot \sin(A) \text{ and } a = CD \cdot \sin(B)}$[/tex].

So, the correct statement from the provided options to complete the proof is:

C. [tex]$\quad b = CD \sin (A)$[/tex] and [tex]$a = CD \sin (B)$[/tex].

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