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

Can this actually be calculated? Unknown angle in a circle inside a right angled triangle
MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : www.youtube.com/@CalculusBoos...
Math Hunter : www.youtube.com/@mathshunter/...
--------------------------------------------------------------------------------
Become a member of this channel to get access of ULTIMATE MATH COURSE
(New video will be uploaded every week)
Join the channel to become a member
/ @mathbooster

[@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°.

[@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.

[@soli9mana-soli4953]

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

[@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.

[@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.

[@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.

[@user-xj5qz3pt3k]

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

[@supercompooper]

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

[@dickroadnight]

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

[@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.

[@jarikosonen4079]

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

[@padraiggluck2980]

Sure, π/2.

[@someonespadre]

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

[@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°

[@daddykhalil909]

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

[@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.

[@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.

[@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.

[@dacupaprilness6254]

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

[@dacupaprilness6254]

Ohhh, do you mean (AD)^2?

[@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]

😂

[@dickroadnight]

AB = AC, by symmetry about BO, as AP = QC Another opportunity to use my favourite theorem! Let (x =) AP = PQ = QC = 1 (scale) BC = 3/√︎2 = BA (by pythag) In triangles CQB and AQB, Cos CQB = - cos AQB (supplement) (1+BQ^2-4.5)/2*1*BQ = (4.5-BQ^2-2^2)2*2*BQ 2(BQ^2 - 3.5) = (0.5 - BQ^2) 3BQ^2 = 7.5 BQ^2 = 15/6 = 5/2 = 2.5 BQ = 1.5811*************** In triangle QBC, by cos rule, Cos QBC = (2.5 + 4.5 - 1)/(2*BQ*3/√︎2) QBC = 26.565 OBC = 45 OBQ = OBC - QBC = 18.4349 In triangle OBQ, by cos rule Cos OBQ = (4r^2 + 2.5 - r^2)/(2*√︎2r*√︎︎2.5) 0.9486 = (3r^2 + 2.5)/(4.472*r) 4.2472r = 3r^2 + 2.5 3r^2 + 2.5 - 4.2472r = 0 r = 0.707033 (1/√︎2) (by quadratic formula) There are now several ways to calculate θ e.g. Θ/2 = 180 - BOQ In half of triangle PQO θ = 2*asin(0.5/r) θ = 2*asin(0.5/(1/√︎2)) θ = 90 degrees Not quick and easy - but it works... I am 74... and I play this game to try to keep a few of my brain cells functional

[@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

[@samanudawatte8255]

😂😂😂😂😂

[@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.

[@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.

[@T0NYD1CK]

ΔPOQ ~ ΔADQ ~ ΔABC ∴ ∠POQ = ∠ABC = θ