SAT Geometry Problem

Question

Circle O (point O is on the origin) passes through points $ A ({\sqrt 3 }, 1$) and point $B (a, 0),$ where $a$ is positive. The measure of angle AOB can be expressed as $\frac{\pi}{b}$. What is the value of $b$ ?

Answer 1

The point $A$ is distance $4$ from the origin, which tells us that the radius of the circle is $2$. Therefore the point $B$ should be $2$ units from the center, so it must be $(2,0)$. Note it could have also been $(-2,0)$ but they told us $a$ is positive.

Now draw a triangle connecting $A = (\sqrt{3}, 1)$, $B = (2, 0)$ and $O = (0,0)$. Let $\theta$ be the angle $AOB$. Then we see that $\cos\theta = \frac{\sqrt{3}}{2}$ and so $\theta = \frac{\pi}{6}$.

Answer 2

This is a problem that looks complicated when you first read it and then becomes simple on a second or third look. Many SAT questions are of this sort. Hint # 1A: remember that you have to be able to solve it in a minute if you are to finish the test, so there can't be too much depth (unlike the real world and really useful questions). Hint # 1B: On the SAT, always look for the trick.

It will take a lot longer to explain the reasoning of this question than to see the answer.

A circle with center at the origin has equation $x^2 + y^2 = r^2$. This s an important fact to remember. Since we know the circle passes through A$(\sqrt3,1)$ then $(\sqrt3)^2 + 1^2 = r^2$ so $4 = r^2$ and $r = 2$. (Remember $r$ is always positive, no question of signed roots here.)

Since the circle passes through B$(a,0)$ with $a$ positive, we can simply look at the graph, or we can work out $a^2 + 0^2 = r^2$ so $a = r$ (given positive) and $a = 2$, and then we see B = $(2,0).

As is often the case with SAT questions, this one is badly worded and makes unjustified assumptions. If $b$ is any non-zero real number, then any angle can be expressed as $\pi / b$. However here the most likely reading would be that $b$ is intended to be an integer and probably the minimum positive integer. Hint #2: On the SAT assume 90 percent of the time the dumb reading (but always always look for a second option, the negative square root, the other side of the absolute value, etc.)

Now you should have followed the basic seven-word rule of geometry and trig: Draw The Diagram And Look At It. That is also Hint #3 for the SAT, Draw The Diagram And Look At It. I would throw a quick sketch in here but have not yet learned how to insert diagrams in this format, sorry. So on your graph you have B $(2,0)$ and A$(\sqrt3,1)$ on a circle center origin and radius = 2. What angle does it look like? If your sketch is not too inaccurate you should have an idea already. Can we prove it?

Yes. Remember the definition of $ sin(\theta)$. On a circle center origin and radius $r$, with a ray OP where P = $(x,y)$, $sin(/theta) = y/r$ Also for completeness $cos(\theta) = x/r$ and $ tan(\theta) = y/x = $ slope of OP. Here P = A = $(\sqrt3,1)$ so $y = \sqrt3$ and we know $r = 2$ so $ cos(\theta) = |sqrt3 / 2$. This gives us a very well-known angle which you should recognize at sight. However you could use the inverse cosine function to be sure.

We find $\theta = 30$ degrees $= \pi / 6$.

Thus we assume (minimum positive) $b = 6$

Answer 3

A sketch is given below. This SAT question is not well posed, edited it as not upto the mark.

$$ a =2,\, b =6\,$$

Answer 4

Unless I've missed something, the circle is completely irrelevant. The required angle is that between $OA$ and the positive $x$-axis, which given the coordinates of $A$ is $$\arctan\frac1{\sqrt3}=\frac\pi6\ .$$ So $b=6$.