In the figure above, the triangle ABC is inscribed in a semicircle, wi

Question

https://gmatclub.com/forum/download/file.php?id=35523 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 gmat_question.png

shashankism · Accepted Answer

I forgot to attach the diagram.. I have edited the diagram provided by you for the solution..

sashiim20 · Answer

Triangle inscribed in a semicircle will always have one angle as  90^{\circ} .

From the diagram  \angle ACB =  90^{\circ}

Connect OC. 
Given;  \angle CÔA = 120°
 \angle COB +  \angle COA = 180
Therefore  \angle COB =  60^{\circ} 
OB = OC (Both are radius). 
Hence  \angle OCB =  \angle OBC =  60^{\circ}

riangle  ABC is 30:60:90 Triangle. Sides of 30:60:90 triangle are in ratio = x : x \sqrt{3}  : 2x

BC : AC : AB = x : x \sqrt{3}  : 2x

AC = 3 \sqrt{3}

Therefore BC = 3 and AB = 6

AB is diameter of semicircle. Radius is half of diameter. Therefore Radius = 3. 
Answer B...

shashankism · Answer

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.

guireif · Answer

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.

danlew · Answer

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.

dave13 · Answer

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 ?

abhimahna · Answer

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?

ujjwal80 · Answer

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.

dcummins · Answer

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

SShareef · Answer

The question says CÔA = 120 degrees. Can we take Ô to be point O?

Bunuel · Answer

___________________________________
It's obviously a typo. Edited. Thank you!

adityasuresh · Answer

Isosceles triangle in the ratio 30:60:90 sides are in the ratio x:root3x:2x

yannichc · Answer

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?

Bunuel · Answer

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.

achloes · Answer

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.

Attachment:

Screenshot 2023-07-27 at 6.01.29 PM.png [ 193.12 KiB | Viewed 3678 times ]

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

In the figure above, the triangle ABC is inscribed in a semicircle, wi

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In the figure above, the triangle ABC is inscribed in a semicircle, wi

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards