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# Pls Help, Due Tmrow

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1)Triangle ABC is an isosceles right triangle where angle A = 90 degrees. O is the circumcenter of Triangle ABC. What is angle AOB in degrees?

2)Let O and H be the circumcenter and orthocenter of acute triangle ABC, respectively. If angle AHC + angle AOC = 240 degrees, what is angle ABC in degrees?

Feb 8, 2018

#1
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I have been thinking about these two problems for some time, and I think I have finally cracked them! I'll now impart my knowledge to you!

1)

 $$\triangle ABC\text{ is an isosceles right triangle}\\ m\angle A=90^{\circ}\\ O\text{ is the circumcenter}$$ Given information $$m\angle A+m\angle B+m\angle C=180^{\circ}$$ Triangle sum theorem (the sum of the interior angles of a triangle equals 180 degrees) $$90^{\circ}+m\angle B+m\angle C=180^{\circ}$$ Substitution property of equality $$m\angle B+m\angle C=90^\circ$$ Subtraction property of equality $$m\angle B=m\angle C$$ Isosceles Triangle Theorem (The angles opposite congruent sides in an isosceles triangle are congruent) $$m\angle B+m\angle B=90^\circ$$ Substitution property of equality $$2m\angle B=90^\circ$$ Simplify $$m\angle B=45^\circ$$ Division property of equality $$\overline{AO}\cong\overline{BO}$$ Property of Circumcenter (The circumcenter is equidistant from the vertices of the triangle) $$m\angle A=m\angle B$$

Well, this is taking some time...

Anyway, $$\triangle AOB$$ is another isosceles triangle. $$m\angle OAB=m\angle B=45^\circ$$, so the remaining angle, $$\angle BOA=90^\circ$$

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Feb 9, 2018
#2
+2419
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The second one was definitely much harder for me to come up with! Here is a picture that may be useful to reference as I solve:

You will probably notice that I added a circle that inscribes the triangle. At its bost basic level, the circumcenter is a point that is equidistant from all the vertices of the triangle. The circumcenter is also the center of a triangle's circumcircle, which I have illustrated with the diagram above. Of course, we know that the sum of $$\angle AHC$$ and $$\angle AOC$$ is 240 by the given info. Notice that $$m\angle AOC=m\angle AHC$$ because they both intercept the same arc. Therefore, we can solve for the angle of the circumcenter.

 $$m\angle AOC+m\angle AHC=240$$ Since we already established that  $$m\angle AOC=m\angle AHC$$, we can solve for the the central angle of the circle. $$m\angle AOC+m\angle AOC=240$$ $$2m\angle AOC=240$$ $$m\angle AOC=120^\circ$$ There is no need to solve for the angle measure of the orthocenter; it is simply a waster of time.

Notice that $$\angle ABC$$ is an inscribed angle, which means that its measure is equal to half the measure of the central angle.

 $$\frac{1}{2}m\angle AOC=m\angle ABC$$ Use substitution here since we already know the measure of one of the angles. $$\frac{1}{2}*120=m\angle ABC$$ $$m\angle ABC=60^\circ$$
Feb 9, 2018
#3
+111331
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1 )

The circumcenter of a right triangle always  occurs at the midpoint  of the hypotenuse

Thus....by  SSS, triangle BOA  is congruent to triangle  COA

Thus angle  AOB  =  angle AOC

But, by Euclid....a line standing upon a line making adjacent angles equal means that the lines are perpendicular.....thus  angle AOB  = 90°

Feb 9, 2018