Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Circle..I can't figure this one out

Expert replies
by mikebarmy » Thu Aug 04, 2011 8:57 pm
I can't figure this one out...Can someone please explain how to arrive at the answer?

A semicircle is on an xy plane with center O at the origin. Points A and B are on the circle. Point A (in the second quadrant) is at point (-radical(3),1). Point B (in the first quadrant) is at point (s, t). Lines drawn from point A to the origin and point B to the origin form a 90 degree angle. What is the value of S?




The answer is 1 <-

I would appreciate any explanation of this. Thanks!
Join the discussion
Source: — Problem Solving |

by gmatboost » Thu Aug 04, 2011 9:33 pm
Let's add a few more points:
Draw vertical lines down to the x-axis from A and from B.
Call these points C and D.
C is at (-sqrt(3), 0) and D is at (s, 0).
Also, put a point Y somewhere along the y-axis, let's say it's at (0, 5) but this is just for reference.

Triangle ACO is a right triangle with legs of length sqrt(3) and 1. You should recognize this as a 30-60-90 right triangle based on the side ratios. The hypotenuse of a 30-60-90 is twice the shorter leg, so AO has length 2.

Angle COA is 30, because it is opposite the side of length 1. This means angle AOY is 60, since angle COY is 90.

Now, we know angle AOB is 90. This means angle YOB is 30. This in turn means that angle BOD is 60.

That means that triangle BOD is also a 30-60-90 right triangle. Since AO and BO are both radii of the semi-circle, they have the same length, so BO, the hypotenuse of triangle BOD, has length of 2 also.

Since angle BOD is 60, the side opposite that, BD, has a length of sqrt(3) because of the 30-60-90 side ratios. Similarly, since angle OBD is 30, the side opposite that, OD, has a length of 1 because of the 30-60-90 side ratios.

That means point B is [spoiler]at (1, sqrt(3)).[/spoiler]
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.


Also, check out the most useful GMAT Math blog on the internet here.
Join the discussion