Prove the law of sines

Answer 1

Draw and altitude from A. Label it "h".

sin B = h / c

sin C = h / b

c sin B = h

b sin C = h

c sin B = b sin C

c / sin C = b / sin B

Q.E.D.

Answer 2

2s = absinC = acsinB = bcsinA; bsinC = csinB; b/sinB = c/sinC; asinB = bsinA a/sinA = b/sinB.

From here a/sinA = b/sinB = c/sinC

Answer 3

(The instruction below should allow you to construct a triangle for illustration purposes. Here is a link to a page that has the triangles:

https://www.desmos.com/calculator/s7y6dbwjjg

)

Our original triangle is ABC.

The angle B is obtuse.

The angle A and C are acute.

The length of AB is c.

The length of BC is a.

The length of AC is b.

We draw a line perpendicular to AC and going to B. We call it BP. Its length is d.

d / a = sin(C)

d = a sin(C)

d / c = sin(A)

d = c sin(A)

a sin(C) = c sin (A)

sin(C) / c = sin (A) / a

To bring in angle B we draw a line perpendicular to BC and going to A. We call it AP'. Its length is d'. Note that this line falls outside the original triangle.

d' / b = sin (C)

d' = b sin(C)

We set B' to be the angle supplementary to B: B' = π - B.

d' / c = sin (B') = sin (π - B) = sin(B)

d' = c sin (B)

b sin(C) = c sin(B)

sin (C) / c = sin (B) / b

So finally:

sin(A) / a = sin (B) / b = sin (C) / c