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guireif
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9

Lets join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.
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guireif
In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If CÔA = 120° and AC = 3√3, what is the radius of the circle?

(a) (4√3)/5
(b) 3
(c) 2√3
(d) 4
(e) 9

Lest join O to C in the diagram.

Let radius of semicircle = x . So , OA= OB = OC = x
Also angle BOC = 180° - angle COA = 60°
Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°

So triangle BOC is an equilateral triangle . Hence BC = x

Now since ACB is a triangle inside semicircle. So Triangle ACB is right angled triangle.

-> AC^2 + BC^2 = AB^2
-> (3sqrt(3))^2 + x^2 = (2x)^2
-> 3x^2 = 27
-> x^2 = 9
-> x =3

Hence radius = 3.

Answer B.

Since OC = OB
angle BCO = angle COB = (180 - angle BOC)/2 = 60°


That's what I was missing!
You're a genius. Thank you so much.

Best,
Gui.
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I tried solving it, but since I was taking more than 1 minute to figure it, I guessed right answer assuming that it has to be a 30:60:90 triangle.
Only the 30:60:90 triangle has a √3 component x : x√3 : 2x.

So, I guessed the diameter should be 2x=6 and hence the radius 3.
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Hi ujjwal80, thanks for an explanation:) i have a couple of questions :) can you please explan. see below. have a great weekend :)

Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?
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Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?

Hey dave13 ,

Here are the answers:

Based on which rule do we assume that the length of OC equals OA and OB? --> We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB

How can we assume that angle BCO is 60° ? --> Since OC = OA, that means triangle COA is isosceles.

We are given that angle COA = 120. => the sum of angles OAC and OCA = 180-120 = 60.

=> angle OCA = 30 (since triangle COA is isosceles).

Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90.

Therefore, BCO = ACN - OCA = 90-30 = 60

Does that make sense?
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dave13
Question 1.

Based on which rule do we assume that the length of OC equals OA and OB?
OC could be of any length, no?

Question 2. regarding this --- > "Angle BCO = angle COB = (180 - angle BOC)/2 = 60° "

How can we assume that angle BCO is 60° ? it could be any degree… no ?

Hey dave13 ,

Here are the answers:

Based on which rule do we assume that the length of OC equals OA and OB? --> We are not assuming. It is the rule. If the given figure is a semi circle with center as O, all the line segments joining the center and any point on the circle are equal and are also known as radii. Hence, OC = OA = OB

How can we assume that angle BCO is 60° ? --> Since OC = OA, that means triangle COA is isosceles.

We are given that angle COA = 120. => the sum of angles OAC and OCA = 180-120 = 60.

=> angle OCA = 30 (since triangle COA is isosceles).

Now, every triangle drawn on the semi circle is right angled triangle. This angle ACB = 90.

Therefore, BCO = ACN - OCA = 90-30 = 60

Does that make sense?

Thanks abhimahna for the explanation, dave13 refer to triangle concepts https://gmatclub.com/forum/math-triangles-87197.html if you have more doubts feel free to post.
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COA = 120 degrees

Recall that the inscribed angle is equal to 1/2 the central angle it intercepts.

So CBA = 60 degrees and the triangle is a 30-60-90 triangle.

We can then use side ratios to determine that AB = 6 and thus radius = 3
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The question says CÔA = 120 degrees. Can we take Ô to be point O?
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The question says CÔA = 120 degrees. Can we take Ô to be point O?
___________________________________
It's obviously a typo. Edited. Thank you!
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Isosceles triangle in the ratio 30:60:90 sides are in the ratio x:root3x:2x
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Why is CBO= 60?

is it bcos the inscribed angle is equal to 1/2x the central angle it intercepts?

so since COA = 120

CB0 = 60? and we know that angles around a straight line.= 180 therefore COB = 60

since we know ACB is a right angle

therefore we know it is a 30-60-90 triangle?
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yannichc

In the figure above, the triangle ABC is inscribed in a semicircle, with center O. If COA = 120° and AC = 3√3, what is the radius of the circle?

(A) (4√3)/5
(B) 3
(C) 2√3
(D) 4
(E) 9


Why is CBO= 60?

is it bcos the inscribed angle is equal to 1/2x the central angle it intercepts?

so since COA = 120

CB0 = 60? and we know that angles around a straight line.= 180 therefore COB = 60

since we know ACB is a right angle

therefore we know it is a 30-60-90 triangle?

    An inscribed angle is exactly half the corresponding central angle.


In this case, angle ABC is half of angle COA, making it 60 degrees.


    A right triangle that is inscribed in a circle must have its hypotenuse as the diameter of the circle. Conversely, if the diameter of a circle is one of the sides of an inscribed triangle, then the triangle must be a right triangle.


Since AB is the diameter of the circle, we know that angle ACB must be 90 degrees. This means that the triangle is a 30-60-90 triangle, which is a "standard" triangle that is important to recognize. The sides of this triangle are always in the ratio 1:√3:2. It's worth noting that the smallest side (1) is opposite the smallest angle (30°), while the longest side (2) is opposite the largest angle (90°).

AC, which is 3√3, is opposite 60°, thus:

AC/AB = √3/2
3√3/AB = √3/2
AB = 6.

Since AB is the diameter, the radius is half of that, making it 3.

Answer: B.
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Angles COA and CBA intercept the same arc AC, therefore the inscribed angle CBA will be 1/2 the central angle COA.

The angles may or may not share a side. Some examples below.

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