As peripheral angles over the same chord, you can equate the two terms: 2x - 1 = 5y + 2 As a center angle, the angle AOB is twice as large, so: AOB = 4x - 2 = 10y + 4 Since the triangle ABO is an isosceles triangle, the angles BAO and OBA are equal. You can conclude 180° - 2 * 63° = 54° for AOB. So: AOB = 4x - 2 = 10y + 4 = 54° 4x - 2 = 54 4x = 56 x = 14 10y + 4 = 54 10y = 50 y = 5 x + y = 19
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Good explanation but these maths are very simple. Please post some tricky and tough math problems.
As peripheral angles over the same chord, you can equate the two terms:
2x - 1 = 5y + 2
As a center angle, the angle AOB is twice as large, so:
AOB = 4x - 2 = 10y + 4
Since the triangle ABO is an isosceles triangle, the angles BAO and OBA are equal. You can conclude 180° - 2 * 63° = 54° for AOB.
So: AOB = 4x - 2 = 10y + 4 = 54°
4x - 2 = 54
4x = 56
x = 14
10y + 4 = 54
10y = 50
y = 5
x + y = 19
Excellent!
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Do you need more circle question? I have tons of question related to circles..
Great explanation👍
Thanks for sharing😊
Only because you have already taught, and that's much, was I able to solve this in seconds, in fact the solution almost jumped off the page.
thank you
An in-your-head quickie, but good fun.
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Thanks for video.Good luck sir!!!!!!!!!!!!
So nice of you
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Please can u tell me the name of the logiciel on which you draw the geometric figures. Good luck
Very nice 👍🙂
x+y=19°
Easy!
It became easy after recalling the Central Angle Theorem
Angle AOB = 180 - 2*63 = 54 = 2*(2X -1) So X = 14.
5Y + 2 = 180 - (90+63) = 27. So Y = 5. ( OA is perpendicular to DB )
X + Y = 19.
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How did you got to know that OA is perpendicular to BD . Please explain.
2x-1=5y+2=(180-2x63)/2=27, x=28/2=14, y=25/5=5, x+y=14+5=19, done.😃
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4x-2=180-126=54
4x=56
x=14
10y+4=54
10y=50
y=5
x+y=14+5=19°
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So nice of you, Pranav
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It is a verbal question
2x-1=5y+2=54/2=27
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But angle in sameline segment are equal so why
19
ACB = 27
ADB=27
X=14
Y=5
X + Y= 19.
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x=14
y=5
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Solved in seconds..
Excellent!
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@@PreMath Thanks bro