Can this actually be calculated? Unknown angle in a circle inside a right angled triangle

[@hcgreier6037]

There is a purely argumentative explanation here without all the math mumbo-jumbo: Since the circle goes through P and Q, the midpoint O must lie on the bisection of PQ. If this circle must *also* touch the triangle in points D and E, the only possibility is that the legs of the triangle AB/BC must have same length. Therefore the bisection of PQ goes through vertex B and makes an angle of 45° (AC has -45°, so to speek). And since P/Q have the same distance from bisection cutting with AC (=midpoint of AC), the angle θ = 90°.

[@jimlocke9320]

At about 11:20, we have ΔPOQ which has 2 sides of length r and one side of length r√2. We should recognize that the side ratio matches that of an isosceles right triangle, so, since Θ is opposite the long side, Θ = 90°. If we miss that and proceed with law of cosines, we could note that the numerator is a² + b² - c². When we do the calculations, the result is a numerator of 0. Therefore, a² + b² = c², which is the Pythagorean theorem. So, ΔPOQ is a right triangle and c = PQ is the hypotenuse. Since Θ is opposite the hypotenuse, Θ = 90°.

[@juanalfaro7522]

No matter what the value of X and R are (we know both > 0), we found EC = AD. Therefore AB = BC (since both sides equal R + the tangent). Thus BCA = BAC = 45. Since DQ is parallel to BC and PE is parallel to AB, OPQ = OQP = 45. Thus, Theta = 90. Thus the calculation is not needed at all, and superfluous.

[@PS-mh8ts]

Nowhere does it say that DQ is parallel to BC. Nowhere is it even implied that D-O-Q are collinear. All that the problem states is that D is the point of tangency and that O is the center of the circle. In fact, if you can prove that DQ is parallel to BC (or that D-O-Q are collinear), that would be another way to arrive at the angle in question.

[@juanalfaro7522]

@@PS-mh8ts At 10:03 it was proven that X = R*sqrt (2). So we can see that POQ is a 45-90-45 triangle since PO = OQ = R. I see now that this is unique to this problem because PQ=X; i.e., if PQ had a different value, say 3X, then we can only say that POQ is an isosceles but not a right triangle. So the discussion could have ended right after 10:03 but was continued because this is usually not the case.

[@juanalfaro7522]

I see now that Theta = 90 only because PQ=X and otherwise, PO and OQ would not be parallel to AB and BC respectively. So the calculation is not superfluous though in this case, on 10:03 it could have been concluded that POQ is a 45-90-45 triangle.

[@graehamebasjan7252]

Why didn't you stop after calculating the angles of the square DBCO, and then stating that angle DOC is equal to theta since they are vertically opposite angles, therefore theta is equal to ninety degrees?

[@ryaj2356]

That’s what I was getting too.

[@tvfamily109]

Because these perpendiculars are not extensions of OP and OQ. At least it’s not how they were drawn

[@graehamebasjan7252]

Look at the given information. It states specifically that O is the centre of the circle... That also implies that OQ is a radius of the circle... Am I mistaken here?

[@jojo_ranjan]

@@graehamebasjan7252Even if OQ is the radius that doesn’t necessarily mean OQ is extension of DO. A line drawn from centre to Any point left or right of Q would also be radius but that doesn’t mean it would be extension of DO.

[@ManjulaMathew-wb3zn]

I noticed some comments assuming D,O and Q to be co linear.Actually they are and you have to prove it before using it. The large triangle is an isosceles triangle so the corner angles are 45 degrees each. Once you figured out r=xsqrt2 you can drop a perpendicular from Q. The height of that perpendicular is calculated to be xsqrt2 which is r. Now this proves those 3 points are co linear. So do the corresponding vertical points. Now you can determine the value of theta to be 90 degrees without calculating.

[@mikefochtman7164]

Once we have triangle OPQ with sides r, r and sqrt(2) r, we see that it fits Pythagoras perfectly (r^2+r^2) = (sqrt(2)r)^2, so it must be a right triangle.

[@pk2712]

This is an interesting problem . Think about the way the problem is set up . We have that triangle ABC is an isosceles right triangle . Then we have a circle that is tangent to sides AB and BC . We then have the hypotenuse intersects the circle at points P and Q . Relax the idea that AP=PQ=QC . That is let the size of the circle vary in size such that PQ=0 . In this case , the circle is tangent to all three sides of the triangle and theta=0degrees . We can increase the size of the circle until PQ=sqrt(2)r . Then AP=QC=0 , and the circle is tangent to the triangle at points A and C . We also have theta=270degrees . We thus have that theta can vary from 0 to 270 degrees . We have from symmetry that AP=QC . The solution to this problem shows that when AP=PQ=QC ; then, theta=90degrees .

[@امیرعبدی-ع3غ]

راه ساده تری هم وجود داشت از آنجایی که peوqd قطر دایره هستند و در مرکز همدیگر را قطع کردند پس دو زاویه o و تتا متقابل‌ به راس بوده و باهم برابرند و چون زاویه o 90 درجه است پس تا نیز 90 درجه میباشد

[@jarikosonen4079]

Is there any other "interesting" cases, where the PQx? (for example if the PQ=2x...)

[@soli9mana-soli4953]

Thanks for cosine formula, I never seen an application like this

[@honghong1977]

ODBE is a quadrilateral, the interior angles sum is 360° The 4th angle must be 90° as the other 3 interior angles are 90°

[@honghong1977]

Using vertically opposite angle principle theta is 90°

[@dantallman5345]

Learned something new and useful at 3:30- 4:00. I did not know that PT^2=PA*PB….the bit about when you have a tangent and chord originating at a common point. Will try to derive this or failing that, look for it. It seems to me that symmetry demands that ABC is isosceles with 45 degree angles. That is intuition, it was nice to see the proof.

[@wusindy8461]

Make simple things complicated! On the 1:40 image. Connect the EQ and get :SAS：RT△PQE≌RT△CQE.=>∠EPQ+PEQ=90°，∠EPQ+∠CEQ=90°,∠EPC+∠C=90°，=> ∠CEQ=∠C=∠EPC=∠PQO=45°，=>∠POQ=90° 回复

[@someonespadre]

If r=1 then AC=3*sqrt2 and the other 2 sides=3

[@supercompooper]

I love the adorable way that you write 'r' 😊

[@dickroadnight]

What he writes as an “r” is actually the geek letter gamma.

[@padraiggluck2980]

Sure, π/2.

[@Ron_DeForest]

I’m curious. Why not just say since the three segments in AC are equal then lines OP and OQ are also equal. P and Q are the intersection of the circle and triangle both are the ratius. There for angle theta must be 90 degrees. Way easier.

[@gibbogle]

Do we need to prove that AB = BC? It seems obvious by symmetry.

[@dickroadnight]

Yes - it seems obvious, but, if it is not given, we should not assume it.

[@gibbogle]

@@dickroadnight The symmetry is clear, it is given by the geometry. This doesn't really help, though, because you still need to find an expression for the length of AB.

[@daddykhalil909]

How can a square have angles??? All 4 angles are right angles. Did I miss something here???

[@T0NYD1CK]

ΔPOQ ~ ΔADQ ~ ΔABC ∴ ∠POQ = ∠ABC = θ

[@dacupaprilness6254]

AD^2=AD? Why is it not sqrtAD?

[@dacupaprilness6254]

Ohhh, do you mean (AD)^2?

[@agronpone7210]

I considerated the triangles ADQ and POQ, their are similarar for the SAS criteria, for which I found AD=2r-> AB=3r; same thing with the triangles PEC and POQ founding that CE=2r-> BC=3r and calling AC= 3x, I found x=r square root of 2 and plus the angle POQ=90 degrees or more simply the traingles ABC and POQ are similar for which the angle POQ=90 degrees.

[@jayeshkumar3861]

Triangle POQ is similar to Triangle DOE. By side r - side r - theta... (Means congruent actually) So angle POQ is 90...(Obviously BDOE is a rectangle (and square) because of Tangent-Normal....

[@siddidandgunnushowleftyt5594]

no it is not given that QD is a straight line i think rest not sure

[@PS-mh8ts]

but nowhere in the problem statement does it say that the points P-O-E or the points D-O-Q lie on a straight line. thus you cannot say that POQ and DOE are opposite angles and hence equal.

[@jayeshkumar3861]

Yes understood​@@siddidandgunnushowleftyt5594

[@jayeshkumar3861]

​@@PS-mh8ts Ya ya.... That's true. r & r are there, not angle. So can't say congruency.

[@siddidandgunnushowleftyt5594]

yes that is what i told@@PS-mh8ts

[@nejirehado7065]

Can't we immediately say that triangle ABC is similar to triangle POQ immediately by SSS postulate upon knowing that all sides of ABC is 3r and all sides of POQ is r? Hence, theta = angle ABC = 90°.

[@MathBooster]

You have to prove that the ratio of sides are equal.

[@nejirehado7065]

Oh, right. PQ ≠ r and AC ≠ 3r. But the ending is still similarity by SSS postulate. Thanks for clarifying.

[@bobansell6041]

No, we can not say this immediately. It assumes theta = 90 degrees. We need to prove this first.

[@johanneschristopherstahle3395]

We actually can: (1) Make a square of side length 3r (2) Divide it into 3x3 little squares with side length r (3) any diagonal inside of one of these must have side length x (4) draw in the circle. You will find that point O has to be on the bottom left of the middle square, hence θ has to be 90°

[@felixv8728]

😂

[@samanudawatte8255]

😂😂😂😂😂

[@heltongama9952]

Since angle tteta is opposite to angle DOE(90º), eta is equal to 90º.

[@gibbogle]

Only true if O is on the line DQ. This you must prove.

[@ECO473]

Where is this relevant in non-STEM working life???

[@Ron_DeForest]

If you aren’t curious, why are you watching?

[@ECO473]

@@Ron_DeForest That is a fair question, sir. This only reminded me of why I failed geometry & trigonometry at my s**thole of a high school... twice...and why I haven't used it in forty-plus years of professional life since. And the teachers there could never tell me where I would use this in real life. What a waste...THEN AND NOW.

[@ishanthagarwal]

Angle DOE and angle POQ are vertically opposite angles so they must be equal and BEOD is square means both angles are 90, why to do all this 🤔🙄

[@MathBooster]

You need to prove that DQ and PE are straight lines. Then only they will be vertically opposite angles.