Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?
I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.
Thank you
Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?
I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.
Thank you
This is a special case of Inscribed and Central Angles in a Circle from Interactive Mathematics Miscellany and Puzzles
When you read the proof keep in mind that your central angle is 180°.
If it should be proven by the Law of Cosines, then,
let yhe two legs of the inscribed triangle be x and y
the hypotenuse is diameter d.
we assume the angle between x and r is 90 degrees.....this angle is opposite d.
By Law of Cosines,
d^2 = x^2 +y^2 -2(x)(y)cos(90deg)
d^2 = x^2 +y^2 -2xy(0)
d^2 = x^2 +y^2 -------------this is the Pythagorean Theorem, applicable to right triangles only. So the inscribed triangle is really a right triangle, whatever the lengths of x and y are.
It sometimes helps to see a dynamic diagram.
There should be a java applet to the right of this text (it might take a second to load)
You can move the red dots around.
Notice that angle C is always a right angle.
On a side note to others: How many knew that you could type in a java applet? I used to use it all the time, but I haven't been here in like a year.
Hello, zodiacbrave!
Well, here's one way . . . my way.
We have a circle with center $\displaystyle O$, diameter $\displaystyle AB$, and radii $\displaystyle OA = OB = OC = r.$Why is a triangle inscribed in a semicircle is always a right triangle?
Draw chords $\displaystyle AC = x$ and $\displaystyle BC = y.$
Let $\displaystyle \angle AOC = \theta,\;\;\angle BOC = \theta'$Code:* * * * * C * o * r * * * * θ * θ' * A o - - - - o - - - - o B * r O r * * * * * * * * * *
Use the Law of Cosines in $\displaystyle \Delta AOC\!:\;\;x^2 \:=\:r^2 + r^2 -2(r)(r)\cos\theta $
. . We have: .$\displaystyle x^2\;=\;2r^2 - 2r^2\cos\theta$ .[1]
Use the Law of Cosines in $\displaystyle \Delta BOC\!:\;\;y^2 \;=\;r^2 + r^2 - 2(r)(r)\cos\theta'$
. . We have: .$\displaystyle y^2 \;=\;2r^2 - 2r^2\cos\theta' $ .[2]
Since $\displaystyle \theta + \theta' \:=\:180^o\quad\Rightarrow\quad \theta' \:=\:180^o - \theta$
. . $\displaystyle \cos\theta' \:=\:\cos(180^o-\theta) \:=\:-\cos\theta $
Substitute into [2]: .$\displaystyle y^2 \;=\;2r^2 - 2r^2(\text{-}\cos\theta)$
. . and we have: .$\displaystyle y^2 \;=\;2r^2 + 2r^2\cos\theta $ .[3]
Add [1] and [3]: .$\displaystyle x^2 + y^2 \;=\;(2r^2 - 2r^2\cos\theta) +(2r^2+2r^2\cos\theta) $
. . $\displaystyle x^2 +y^2 \:=\:4r^2 \quad\Rightarrow\quad x^2 + y^2 \:=\:(2r)^2$ . . . Pythagorus!
Therefore, $\displaystyle \Delta ABC$ is a right triangle with $\displaystyle \angle C = 90^o.$