Complete the proof of the Law of Sines/Cosines.

Triangle ABC with side b between points A and C, side c between points A and B. Segment drawn from point A to point D where D is between points B and C, segment AD is labeled x.

Given triangle ABC with altitude segment AD labeled x. Angles ADB and CDA are _1._ by the definition of altitudes, making triangle ABD and triangle CDA right triangles. Using the trigonometric ratios sine of B equals x over c and sine of C equals x over b. Multiplying to isolate x in both equations gives x = _2._ and x = b ⋅ sinC. We also know that x = x by the reflexive property. By the substitution property, _3._. Dividing each side of the equation by bc gives: sine of B over b equals sine of C over c.
A)
1. altitudes
2. b ⋅ sinB
3. b ⋅ sinB = c ⋅ sinC
B)
1. right angles
2. b ⋅ sinB
3. b ⋅ sinB =c ⋅ sinB
C)
1. altitudes
2. c ⋅ sinB
3. c ⋅ sinB = b ⋅ sinC
D)
1. right angles
2. c ⋅ sinB
3. c ⋅ sinB = b ⋅ sinC

Option D)

Explanation:

After completing the text it looks like this:

Given triangle ABC with altitude segment AD labeled x.

Angles ADB and CDA are     right angle      by the definition of altitudes, making triangle ABD and triangle CDA right triangles. Using the trigonometric ratios sine of B equals x over c and sine of C equals x over b. Multiplying to isolate x in both equations gives      x = c . sinB       and x = b ⋅ sinC. We also know that x = x by the reflexive property. By the substitution property,       c . sinB = b . sin C     . Dividing each side of the equation by bc gives: sine of B over b equals sine of C over c.

1. right angle

The altitude of a triangle is defined as the line segment that goes perpendicular from one a side (or the extension of a side) to the opposite vertex.

Thus the line segment x (given as the altitude) runs perpendicular from the vertex A to the point D on the segment BC, which makes both the angle ADB and ADC 90º or right angles.

2. x = c. sinB

The trigonometric sine ratio is defined for right angles as the opposite leg divided by the hypotenuse.

Focusing In the triangle ADB, the opposite leg to the angle B is x and the hypotenuse is c.

Therefore, sinB = opposite leg / hipotenuse or:

From that you can solve for x, multiplying both sides by c, which yields:

Focusing In the triangle ADC, the opposite leg to the angle C is x and the hypotenuse is b.

Therefore, sinC = opposite leg / hipotenuse or:

3.  c ⋅ sinB = b ⋅ sinC

From that you can solve for x, multiplying both sides by b, which yields:

Now, since x = x, you substitute and get:

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)