Calculate the distance AB | Important Geometry and Algebra skills explained

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Calculate the distance AB | Important Geometry and Algebra skills explained

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[kennethstevenson]

Great step by step illustration, the best on the internet, easy to follow, complete and correct.

[bobjordan]

These tutorials are VERY well done! I solve a goodly amount of them (not all). I'm a licensed professional engineer and if students had explanations like this back in the day a lot more people would have finished their engineering degrees.

[-wx--]

Area of △ABO = ½AO·28 = ½AB·35 ⇒ AO = 5⁄4AB. Right triangle with 5:4 hypotenuse-to-leg ratio is 3:4:5 triangle, thus AB = 4⁄3BO.

[Copernicusfreud]

I used the Pythagorean theorem to solve the sides of triangle POC as 21, 28 and 35. I then used sin (angle PCO) = opposite/hypotenuse. sin (angle PCO) = 21/35 = 0.6, and (angle PCO) = 36.869898. (angle PCO) = (angle CAP). sin (angle CAP) also equals 0.6. Using the law of sins, c/sin 90 = 28 / sin (angle CAP) or c/1 = 28/0.6 or c = 28/(3/5) = 140/3 or approximately

[philipkudrna]

I hadn‘t seen the like triangles and used trigonometry to calculate the angles and the the side AB and arrived at 46, 667 as well. But the like-triangle proportion approach is much more elegant and easier to calculate!

[montynorth]

If triangle OPB is recognised as a 3, 4, 5, then length OP is quickly solved as 21, each side being multiples of 7.
Triangles OPB and OAB are similar. (just different choice of larger triangle)
AB / 35 = BP / OP.
AB / 35 = 28 / 21.
AB = 28 x 35 / 21.
AB = 4 x 35 /3.
AB = 140/3 = 46.67.

[Mike-evyn]

Cut in half the triangle(OBC) inside the circle into a triangle with one side = 28 (half of 56) and a hypotenuse of 35. Solve for the other side using pythagorous = 21. Now using the Law of sines find the two angles (angle = opp side/hypotenuse. The two angles compute to 53.130102 and 36.869898. The third angle is known = 90. Now solve for the hypotenuse of the big triangle formed by ABO = AB = The square root of 53.130102 - the square root of 35 = to infinity. I am 73 years old and still love this stuff.

[tombufford]

At a quick glance: Two tangents giving right angles and equidistant from an external point to circle tangents. Then midpoint of BC, D = 0.5 * BC = 28. Angle OBD = cos^-1 (28/35) = 36.87 degrees. Angle DBA =90 - 36.87 = 53.13 . Angle DBA = 90-53.13 = 36.87 . Distance AB = 35/(Tan(36.87)) = 46.67.

[towguy]

56/2=28
triangle Opb is right triangle measuring x, 28, 35
similar triangle is 1/7x, 4, 5
triangle Opb is a 3, 4, 5 right triangle. with measurements of 21, 28, 35

Angle cab is 2* angle cOb
Angle pbO=angle pab because similar triangles
Line bo is similar to line ab
looking at 3, 4, 5 ratio, line AB=line PB/3*5=28/3*5=46.667

[theoyanto]

Great example once again, before watch your videos I had no idea geometry had so much going for it, love it

[timeonly]

Once you've established that triangle ABO is a right triangle, then use the fact that dropping a perpendicular from its right angle to the hypotenuse give you 3 similar triangles: ABO, APB & BPO.

[egillandersson]

I get it using the law of sines :
- put α = ∠ABC,
- find the other angles : ∠BAC = 180°-2α, ∠BOC = 2α, ∠OBC = 90°-α
- apply the law of sines in triangle OBC and you get sin α = 4/5 and then cos α = 3/5
- apply the same law in triangle ABC

[philipellis]

Once you recognize that the triangles are 3-4-5 rt. triangles, then the set up is AB = 4/3 x 35 ! No need to use the Pythagorean Theorem, just a simple ratio.

[seriouspossibilistaboutx]

(Using area of kite), Let AB = c; therefore 35x = (35^2 + x^2)^0.5 * 56; x = 140/3 is approx 46.66667

[HappyFamilyOnline]

Awesome presentation👍
Thanks for sharing😊

[alittax]

I love your work, thank you very much! Please keep making these videos, I'm learning a lot!

[sergeyvinns]

Triangle BPO OB=5*7, BP=4*7, PO=3*7. Triangle ABO OB=3*35/3=35, AO=5*35/3=175/3, AB=4*35/3=140/3=47, 6(6).

[alexundre]

Good Morning Master
A Hug From in Brasil 🇧🇷
Thank so much for classes

[vara]

It took a lot of theorems and calculations to find the length of the tangent. Has this problem useful in any practical situation? E.g. the distance of a circular object from a point?

[quigonkenny]

Let D be the point where OA intersects BC. As CA and BA are tangents, ∠OBA and ∠OCA are 90° angles, and by Two Tangent Theorem, BA = CA. Therefore ∆OBA and ∆OCA are congruent. By observation, OB = 35 and BD = 56/2 = 28, so ∆OBD is a 7:1 ratio (3, 4, 5) Pythagorean Triple triangle and thus OD = 3(7) = 21. Let ∠BOD = α and ∠DBO = β. By complementary angles, ∠ABD also = α and thus ∠DAB also = β. Thus ∆ODB and ∆BDA are similar.

AB/BO = DB/OD
AB/35 = 28/21
AB = 28(35)/21 = 4(35)/3 = 140/3 = 46⅔