Question

Pleaseee Helpp!!!

Answer 1

Answer:

Given: In ΔABC , $AD \perp BC$

To prove that: $\frac{\sin B}{b} =\frac{\sin C}{c}$

$AD \perp BC$           [Given]

In ΔADB

The sine angle is defined in the context of a right triangle is the ratio of the length of the side that is opposite that angle to the length of the longest side of the triangle.

$\sin B =\frac{h}{c}$             [By definition of sine]                 .....[1]

Multiplication Property of equality states that you multiply both sides of an equation by the same number.

Multiply by c to both sides of an equation [1] we get;

$c \cdot \sin B =c \cdot\frac{h}{c}$

Simplify:

$c \sin B = h$                    ......[2]

Now, In ΔACD

Using definition of sine:

$\sin C =\frac{h}{b}$

Multiply both sides of an equation by b;

$b \cdot \sin C = b \cdot \frac{h}{b}$       [Multiplication Property of equality]

Simplify:

$b \sin C = h$                     ......[3]

Substitute [3] in [2];

$c \sin B = b \sin C$              ......[4]

Division property of equality  states that if you divide both sides of an equation by the same nonzero number the sides remains equal.

[4] ⇒$\frac{\sin B}{b} =\frac{\sin C}{c}$

Therefore, the missing statement in step 6 is; $c \sin B = b \sin C$    

Answer 2

In step 6 we substitute the value we just found for h in step 5 into the equation in step 3. That's also an equation for h, so we're just setting the other sides equal:

$c \sin B = b \sin C$

Third choice.