Can you find the angles X and Y? | Justify your answer

Can you find the angles X and Y? | Justify your answer | (Step-by-step explanation) |

Рет қаралды 5,742

PreMath

Күн бұрын

Пікірлер: 34

@MrMichelX3 10 ай бұрын

I like your videos much because I always learn new theorems.

@PreMath 10 ай бұрын

Glad to hear it! Thanks a lot ❤️

@arnavkange1487 10 ай бұрын

Can u give the proof for alternate segment theorem in next vedio ...as it is little bit confusing

@19-biswarooptalukdar99 10 ай бұрын

Yes... Please it is very much confusing....

@solcubing 10 ай бұрын

I advise you to get a pen and paper so tracking the angles are easier. Let ΔABC be in a circle with centre O, such that A, B and C touch the circumference of this circle. The tangent DE passes through A on the circumference of the circle. AB is a chord where C lies on the major arc. By drawing OA and OB, two new radii are constructed (these lines go from the centre to the circumference of the circle so they are radii). This means that OA and DE are perpendicular (the tangent to a circle meets the radius of that point at 90°). As OA and OB are radii, ΔOAB is isosceles. This also means that angles OAB and OBA are equal; these should be labelled x. Draw the radius that is perpendicular to chord AB. Label the intersection point F (Note that FA=FB because a radius that is perpendicular to a chord bisects that chord). ΔOFA and ΔOFB are congruent, therefore, angles AOF and BOF are also the same; these should be labelled y. Since ΔOFA is a right angle triangle, x+y=90°. This means that angle FAD=y (90-x=y). Angle AOB=2y (because originally AOB was made up of angles AOF and BOF which equaled y+y). As the angle at the circumference is twice the angle at the centre, we can state that angle AOC is twice the size of angle ACB, so ACB=y, resulting in angle DAB=angle ACB. As AB is the chord, angle ACB is in the alternate segment to angle DAB and so angles in the alternate segment are equal.

@PreMath 10 ай бұрын

Sure! Pretty soon. Thanks ❤️

@innovativeeducation5814 10 ай бұрын

a + x = 90 , (radius subtends 90 at point of contact of tangent) y + a + a = 180 ( angle sum proprty of triangle) y + 2a = 2 ( x + a ) ( from above two statements) y + 2a = 2x + 2a y = 2x x = y /2............ Hence Proved

@arnavkange1487 10 ай бұрын

Thanks solcubing and innotiveeducation for giving me the proof

@HappyFamilyOnline 10 ай бұрын

Great presentation 👍 Thank you so much for your hard work 😊

@PreMath 10 ай бұрын

My pleasure 😊

@Ibrahimfamilyvlog2097l 10 ай бұрын

Nice sharing 🌹🌹

@PreMath 10 ай бұрын

Thank you! Cheers! ❤️

@bigm383 10 ай бұрын

Fascinating!❤😀🥂

@PreMath 10 ай бұрын

🌹 Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@engralsaffar 10 ай бұрын

y=360-2(2x-15)=390-4x In the equilateral triangle y+2a=180 Tangent is perpendicular to the radius x+a=90 2x+2a=180 Therefore, y=2x 2x=390-4x 6x=390 x=65 y=130

@PreMath 10 ай бұрын

Excellent! ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@murdock5537 10 ай бұрын

Nice! φ = 30°; OPQ = x + z = 3φ → z = OPB = PBO = 3φ - x → 2z = 6φ - 2x → y + 2z = 6φ = y + (6φ - 2x) → y - 2x = 0 → y/2 = x → BOP = y → BAP = y/2 → BAP + PCB = 6φ = y/2 + (2x - 15°) = x + 2x - 15° = → 3x = 6φ + 15° → x = (6φ + 15°)/3 = 2φ + 5° = 65°

@PreMath 10 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍