in the figure ,O is the centre of the circle .if length AC is 33 length CB is 44 then lenght Co is?
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Answered by
113
Hello Mate!
Since AB is straight line therefore by theorum < ACB is 90°.
So, ∆ABC is a right triangle.
Pythagoras theorum
H² = B² + P²
AB² = AC² + BC²
AB² = 33² + 44²
AB = √( 1089 + 1936 )
AB = 55 cm.
Since AB is a diameter then,
AB = 2AO
55 cm = 2 × AO
27.5 cm = AO
Since, AO = CO = BO = radius.
Hence CO is 27.5 cm.
Have great future ahead!
Since AB is straight line therefore by theorum < ACB is 90°.
So, ∆ABC is a right triangle.
Pythagoras theorum
H² = B² + P²
AB² = AC² + BC²
AB² = 33² + 44²
AB = √( 1089 + 1936 )
AB = 55 cm.
Since AB is a diameter then,
AB = 2AO
55 cm = 2 × AO
27.5 cm = AO
Since, AO = CO = BO = radius.
Hence CO is 27.5 cm.
Have great future ahead!
Priyanka these questions are usually asked after theorems. If we are well versed with the applications of theorems, then we can play with sums :)
Theorem used here is: Angle inscribed in a semicircle is 90 degrees
yes
osm
Thanks Dmohit and Sakshi sis. Thanks Kalpesh bhaiya for support!
No mention :)
Thanks Bijaymourya
Answered by
58
AO= OC= OB= r = radius
As we know diameter subtends right angle at any point on circumference
therefore <C= 90
In Triangle ACB
let <CAB= a
So tan a= BC/AC= 44/33= 4/3
As we know tan53 = 4/3
So a becomes 53 degree
Now In triangle ACB
Sin 53 = BC/AB = 44/2r
As we know sin 53 = 4/5
So 44/2r = 4/5
44× 5= 2r× 4
220= 8r
r= 220/8= 27.5
So length of CO = r = 27.5
✌✌✌Dhruv✌✌✌✌✌✌
As we know diameter subtends right angle at any point on circumference
therefore <C= 90
In Triangle ACB
let <CAB= a
So tan a= BC/AC= 44/33= 4/3
As we know tan53 = 4/3
So a becomes 53 degree
Now In triangle ACB
Sin 53 = BC/AB = 44/2r
As we know sin 53 = 4/5
So 44/2r = 4/5
44× 5= 2r× 4
220= 8r
r= 220/8= 27.5
So length of CO = r = 27.5
✌✌✌Dhruv✌✌✌✌✌✌
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Great answer :)
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