In these two right triangles, ∠A and ∠B are both opposite angles to h. Therefore, by applying the definition of the sine ratio to ∠A and ∠B, it is possible to relate the sine of these angles with the side lengths a and b.
Next, h can be isolated and written in terms of the corresponding side length and angle, for both right triangles.
By following the same procedure but drawing the height from vertex A, it can be shown that bsinB=csinC. Putting these two results together, the Law of Sines is obtained.
This proof is very similar to the proof for acute triangles, but it uses an interior and an exterior height. First, the height h1 from the vertex where the obtuse angle is located will be drawn. Just as before, this generates two right triangles.
The sine ratio will be written for these right triangles.
In the equations, h1 can be isolated.
sinA=ch1sinC=ah1⇔h1=csinA⇔h1=asinC
By the Transitive Property of Equality, it can be stated that csinA and asinC are equal.
{h1=csinAh1=asinC⇒csinA=asinC
The obtained equation can be rearranged to obtain the desired result.
Next, the exterior height h2 from vertex C will be drawn. Let E be the point of intersection of this height and the extension of AB.
Here, ∠ABC and ∠CBE form a linear pair and are therefore supplementary angles. Because the sine of supplementary angles is the same, the sine of ∠ABC equals the sine of ∠CBE. Also, using the sine ratio on △BCE, it can be stated that the sine of ∠CBE is the ratio of h2 to a.
sinABC=sinCBEandsinCBE=ah2
By the Transitive Property of Equality, the sine of ∠ABC can be written in terms of h2 and a.
⎩⎪⎨⎪⎧sinABC=sinCBEsinCBE=ah2⇓sinABC=ah2
Now △ACE will be considered. By using the sine ratio, it follows that the sine of ∠A is the ratio of h2 to b.
Now, h2 can be written in terms of sinABC and a, and in terms of sinA and b.
sinABC=ah2sinA=bh2⇔h2=asinABC⇔h2=bsinA
The Transitive Property of Equality can be used one more time.
{h2=asinABCh2=bsinA⇓asinABC=bsinA
If only △ABC is considered, then ∠ABC can be named as ∠B.
This allows sinABC to be written as sinB. Therefore, asinB=bsinA, and this equation can be rearranged to obtain the desired formula.
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