A fun little problem I can't solve.

mathvalde said:
View attachment 30443
This is taken from the 2022 Georg Mohr test.
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Have you observed what angle BAC is? Or at least seen a couple similar triangles? (The first thing I did is to draw in the radius to the point of tangency, which is always necessary.)

There are several ways to solve this.
 
Thank you! I completely missed the trigonometry part of this. After drawing the radius to the point of tangency I could figure the angles out, and I got the answer to be C.
 
mathvalde said:
Thank you! I completely missed the trigonometry part of this. After drawing the radius to the point of tangency I could figure the angles out, and I got the answer to be C.
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No trigonometry needed - that is use of sin, tan, cos etc.

Only geometry - circles and similar triangles....
 
mathvalde said:
Wait, how do I solve it by only using geometry? I am curious now.
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Call T , the tangent point on AB. Then

BC/OT = AB/OA = AC/TA

AC = 2* OA and OT = 1 and Pythagorus .........continue......
 
Dr.Peterson said:
Show us your work with trig, and we can probably turn it into geometry. After all, trig ratios are really just ways to name similar triangles (sort of).
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After doing the radius to the point of tangency I could make to known sides in the smaller similar triangle. Side "a" must be 1 and side "c" must be 2. Then I calculated sin^-1(a/c) which gave me 30°. Now I know all the angles since it was a similar triangle. Lastly, I just did TanA*b to find the side BC. It gave 4/√3. Hope this makes sense :)
 
mathvalde said:
After doing the radius to the point of tangency I could make to known sides in the smaller similar triangle. Side "a" must be 1 and side "c" must be 2. Then I calculated sin^-1(a/c) which gave me 30°. Now I know all the angles since it was a similar triangle. Lastly, I just did TanA*b to find the side BC. It gave 4/√3. Hope this makes sense :)
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SK's method in effect uses the Pythagorean theorem to find your tangent of A given the sine of A, without needing to explicitly find angle A.
 
Dr.Peterson said:
SK's method in effect uses the Pythagorean theorem to find your tangent of A given the sine of A, without needing to explicitly find angle A.
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Exactly, and that is also probably the right way to solve it. I'm so relieved haha, I don't why this rather simple problem irritated me so much.
 
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