Question

Complete the proof of the Law of Sines/Cosines.

Answer 1

Corrected Question:

Complete the proof of the Law of Sines/Cosines.  

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 ACD 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.

Answer:

(1) right angles

(2) c sin B

(3) c sin B = b sin C

Step-by-step explanation:

As shown in the diagram attached to this response,

(i) the given triangle is ABC with altitude segment AD labeled as x,

(ii) side AB is labeled as c, side AC is labeled as b, and

(iii) angles ADB and CDA are right angles by definition of altitudes.

(iv) Therefore, triangle ABD and CDA are right-angled triangles.

(v) Now, using trigonometric ratios, sine of B equals x over c. i.e

sin B = $\frac{x}{c}$          --------(i)

(vi) Also, sine of C equals x over b. i.e

sin C = $\frac{x}{b}$         ------------(ii)

(vii) Multiplying to isolate x in both equations gives;

x = c sin B           [from equation (i)]           ------------(iii)

x = b sin C          [from equation (ii)]             -----------(iv)

(viii) We know that x = x by the reflective property. i.e x in equation (iii) is equal to x in equation (iv). By substitution property, this becomes,

c sin B = b sin C           -----------------(v)

(ix) Dividing each side of the equation in equation (v) by bc gives: sine of B over b equals sine of C over c i.e

$\frac{csinB}{bc} = \frac{bsinC}{bc}$

$\frac{sinB}{b} = \frac{sinC}{c}$             [This is the sine rule]