m
Mathematic
High School student
A park maintenance person stands 16 m from a circular monument. If you assume her lines of sight form tangents to the monument and make an angle of 34°, what is the measure of the arc of the monument that her lines of sight intersect?
A. 23.5°
B. 94°
C. 266°
D. 133°
We can start by drawing a diagram of the situation:
[asy]
pair O = (0,0);
draw(Circle(O,1));
pair A = dir(34);
pair B = (-1,0);
draw(O--A--B);
draw(rightanglemark(O,A,B,2));
label("$16$ m", (O+A)/2, NW);
label("$34^\circ$", O, 2*dir(17));
[/asy]
We want to find the angle that the two tangents form at the center of the circle. Since the tangents are both perpendicular to the radius that connects the center of the circle to the point of tangency, we can draw radii from the center of the circle to the points of tangency and create a right triangle:
[asy]
pair O = (0,0);
draw(Circle(O,1));
pair A = dir(34);
pair B = (-1,0);
draw(O--A--B);
draw(rightanglemark(O,A,B,2));
label("$16$ m", (O+A)/2, NW);
pair C = foot(O, A, B);
draw(O--C);
draw(rightanglemark(O,C,A,4));
label("$r$", O--C, S);
[/asy]
Since $\angle AOC$ is a right angle, we have $\sin 34^\circ = \frac{16}{r}$. Solving for $r$, we get $r = \frac{16}{\sin 34^\circ}$.
The measure of the arc that the tangents intersect can be found by considering the central angle that subtends that arc. Since the radius forms a right angle with the tangent at the point of tangency, the central angle $\theta$ that the two tangents form is half of the angle between the tangents:
[asy]
pair O = (0,0);
draw(Circle(O,1));
pair A = dir(34);
pair B = (-1,0);
pair C = foot(O, A, B);
draw(O--A--B);
draw(rightanglemark(O,A,B,2));
draw(O--C);
draw(rightanglemark(O,C,A,4));
label("$16$ m", (O+A)/2, NW);
label("$r$", O--C, S);
label("$\theta$", O, 1.2*dir(8));
[/asy]
Since we have $r$ and $16$, we can use the Pythagorean theorem to find the length of the segment from the center of the circle to the point of intersection of the tangents:
$$\sqrt{r^2 - 16^2} = \sqrt{\left(\frac{16}{\sin 34^\circ}\right)^2 - 16^2} = \frac{16}{\sin 34^\circ}\sqrt{1-\sin^2 34^\circ} = \frac{16\cos 34^\circ}{\sin 34^\circ}.$$ Therefore, we have $\tan \frac{\theta}{2} = \frac{8\cos 34^\circ}{\frac{16}{\sin 34^\circ}} = 2\sin 34^\circ \cos 34^\circ = \sin 68^\circ$. Thus, $\frac{\theta}{2} = \arctan \sin 68^\circ$, so $\theta = 2\arctan \sin 68^\circ$.
Using a calculator, we find that $\theta \approx 133^\circ$, so the answer is $\boxed{\textbf{(D)}\ 133^\circ}$.