Grades 9-10 Video Solutions 2021-video 2021 9-10/15

Video Transcription

Problem number 15 states, the figure shows a semicircle with center O. Two of the angles are given. What is the size in degrees of the angle alpha?  So here we have the diagram the problem gives us. And the problem is going to be quite hard to solve if we don't know or don't figure out a certain theorem.  This theorem is called Thales Theorem, and it involves triangles and semicircles, kind of like what we have here. So formally, this is what it says.  If AC is a diameter, and B is a point on the diameter circle, then the angle ABC is a right angle. So from this theorem, we can show that this is a right angle.  So now let's go back to the original question. Let's consider this triangle. And you might notice that this triangle has two legs that are the radii of a circle.  And since all the radii of a circle are the same, that means that this triangle must be isosceles.  This means that the angles opposing the equal sides will have the same value of 32 degrees. This trick is common with circles and triangles.  So if you're ever faced with circles and triangles in a problem, look for isosceles triangles made up of the radii of a circle.  Turns out there's another isosceles triangle right here also made up of radii. Using the same trick, we can notice that the angle in the top right is 67 degrees.  Now we have a big mess in the top corner that we can express algebraically. This would be that the two angles that we just found, 32 and 67 degrees, together sum  up to be equivalent to that of the right angle and alpha. So now let's solve for alpha. We get 99 equals 90 plus alpha.  Subtract 90 from both sides, and we get that alpha equals 9 degrees. So the question asked us, what is the size in degrees of the angle alpha? The answer is 9 degrees. Letter A.

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