SAT Math IX

BO/CO = cos(180-120) = 0.5 CO = 2 * BO = 2*R CF = 2 * CO = 2 * 2 * R = 4R

Sin30=r/co , co=2r cf=2co then cf=4r

ACDF is rectangle, therefore ∠A, ∠C, ∠D & ∠F is 90 degree ∠COE=120 & ∠OEF= 90 Therefore ∠EOF= 180-120=60 and ∠OFE= 30 degree As, AC||DF and Line CF passes through center O, therefore CO=OF

ΔOEF has three angle of 90, 60 & 30. And OE= r Sin ∠OFE=OE/OF = r/OF OF= r/ Sin30= 2r As OF=CO, Therefore, CF= 2* OF= 4r

The phrase in terms of means that this is a Plugging In question. The figure above the question is not drawn to scale, so redraw or add to the figure based on information in the question.

Because $\overline {AC}$ and $\overline {DF}$ are tangent to the circle, they form a $90°$ angle with a radius or diameter of the circle. Draw in diameter $\overline {BE}$ and mark the right angles. Draw the diagonal of the rectangle $\overline {CF}$ . The question states that $\angle COE=120°$ so $\angle COB=60°$ . We have a 30-60-90 triangle, $COB$ , in which $\overline {OB}$ is the radius of the circle and the shortest side of the triangle.

Time to Plug In! If $r=5$ , the sides of this special right triangle are $5:5\sqrt{3}:10$ . So $CO=10$ , which is half the length of $CF$ , which must be equal 20. Now Plug In $r=5$ in the answers:

$A$ results in $10 \sqrt{3}$ ;

$B$ gives $20$ ;

$C$ gives $20\pi$ ;

$D$ gives $10 \sqrt{2}$ .

Only $B$ has a result of $\boxed{20}$ .

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Solution credit: The Princeton Review

in quadrilateral COED angle CDE =90 , angle OED=90 & angle given that angle COE=120 . thus applying angle sum rule angle OCD=360-120-90-90=60 . now in triangle CFD CD=2R & by angle sum property of triangle angle CFD= 180-90-60=30 . now sin30=2R/CF, so CF=2R/Sin30=2R*2=4R

Angle COE = 120, so angle COB is 60. Using trigonometry where cos COB = BO/CO. Thus, cos 60 = r/CO <=> CO = 2r. CO = OF, hence CF = CO+OF. CF = CO + CO = 2CO = 2(2r) = 4r

Let the midpoint of $\overline{CD}$ be $M$ . Note that since $\angle MOE = 90$ and $\angle COE = 120$ , we have $\angle COM = 30$ and $\angle BOC = 60$ , and thus $\angle OFE = 60$ as well. Now, let the radius be $r$ . By $30-60-90$ triangles, we have $\overline{CO} = \overline{FO} = 2r \rightarrow \overline{CF} = 2r + 2r = \boxed{4r}$ .

You should say ANGLE COE measures 120 DEGREES