Please show us what you've tried, as we ask you to:View attachment 30443
This is taken from the 2022 Georg Mohr test.
No trigonometry needed - that is use of sin, tan, cos etc.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.
Wait, how do I solve it by only using geometry? I am curious now.No trigonometry needed - that is use of sin, tan, cos etc.
Only geometry - circles and similar triangles....
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).Wait, how do I solve it by only using geometry? I am curious now.
Call T , the tangent point on AB. ThenWait, how do I solve it by only using geometry? I am curious now.
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 senseShow 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).
Thank you very much. This helped me understand everything.Call T , the tangent point on AB. Then
BC/OT = AB/OA = AC/TA
AC = 2* OA and OT = 1 and Pythagorus .........continue......
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.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
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.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.
I am so glad that you understood everything.....Thank you very much. This helped me understand everything.