As shown in the figure above, line segments AB and AC are tangent to circle O. If line segments BD and DA have the same length, what is angle BAO? (Note: Figure not drawn to scale.)

A. 15º B. 30º C. 36º D. 45º E. 50º

(B) For geometry problems that are this complicated, you should re-draw the diagram while reading the explanation, and label along with the explanation.

Since line segments AB and AC are tangent to the circle, angles OBA and OCA are right angles. Let x be the degree measure of angle BAO and let y be the degree measure of angle BDA. We will now solve for the other angles in terms of x and y. If we can get two distinct equations with these two variables, we can solve for both x and y. The value for x is the answer to the question.

Since triangle ADB is isosceles (it has two legs with the same length): Angle DBA = Angle BAD = x.

Since the sum of the degree measures of the angles of triangle ADB must be 180º, we can also say that: 2x + y = 180. This gives us one equation involving x and y.

Then, since triangle BOD is isosceles (because OB and OD are both radii and therefore the same length), we know that angle OBD has that same degree measure as angle ODB. We can put angle OBD in terms of x and angle ODB in terms of y, and set these two expressions equal to each other.

Angle OBD = 90º – x Angle ODB = 180º – y.

Now let's set these expressions equal to each other and simplify: 90º – x = 180º – y y – x = 90º.

Now we have two equations and two unknowns: 2x + y = 180 y – x = 90º.

So we can solve for x. The easiest way to solve for x is to simply subtract the second equation from the first, causing the y terms to cancel. This yields: 3x = 90º x = 30º.

The correct answer is choice (B). -------------

How do you get Angle OBD = 90 – x and Angle ODB = 180 – y?

Users browsing this forum: No registered users and 2 guests

You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum

GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.