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

[@MrMichelX3]

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

[@PreMath]

Glad to hear it! Thanks a lot ❤️

[@Ibrahimfamilyvlog2097l]

Nice sharing 🌹🌹

[@PreMath]

Thank you! Cheers! ❤️

[@bigm383]

Fascinating!❤😀🥂

[@PreMath]

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

[@HappyFamilyOnline]

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

[@PreMath]

My pleasure 😊

[@engralsaffar]

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]

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

[@murdock5537]

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]

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

[@wackojacko3962]

Ooooo, I absolutely love it!

[@PreMath]

Thank you! Cheers! ❤️

[@mathbynisharsir5586]

Nice video sir

[@PreMath]

Thanks dear ❤️🌹

[@AmirgabYT2185]

x=65° y=130°

[@giuseppemalaguti435]

(180-y)/2+x=90..y=2x...y=2(180-(2x-15))=390-4x=390-2y...3y=390...y=130,x=65

[@PreMath]

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

[@arnavkange1487]

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

[@19-biswarooptalukdar99]

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

[@solcubing]

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]

Sure! Pretty soon. Thanks ❤️

[@innovativeeducation5814]

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]

Thanks solcubing and innotiveeducation for giving me the proof

[@arnavkange1487]

This was an easy question ...

[@PreMath]

Great! ❤️

[@romangarland47]

Could you guys please suggest me some pure mathematics book?

[@misterenter-iz7rz]

y=2x, x+2x-15=180, 3x=195, x=65, y=130.😊

[@PreMath]

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

[@prossvay8744]

X=65, y=130

[@PreMath]

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

[@johnwindisch1956]

Great I came close just need to remember that 90-65 = 25. Not 15!