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Define trigonometric ratios and solve problems involving right triangles Videos 5 videos

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SAT Math 1.3 Geometry and Measurement
231 Views

SAT Math 1.3 Geometry and Measurement. Find the length of CE.

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SAT Math 1.3 Geometry and Measurement 231 Views


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Description:

SAT Math 1.3 Geometry and Measurement. Find the length of CE.

Language:
English Language

Transcript

00:03

Here’s your shmoop du jour, brought to you by the length of a line.

00:07

Although…we hear it’s been using supplements.

00:10

Find the length of CE.

00:13

Here's a little picture and we're looking for the length of this line right here.

00:18

And here are the potential answers...

00:22

We know right off the bat, for example, that triangle CDE is a 30-60-90 triangle.

00:37

…and we remember that the sides of a 30-60-90

00:40

triangle always have a ratio of x to x square root of 3 to 2x.

00:46

It’s pretty obvious that DE is the longer of the two legs…

00:50

…and even if we weren’t sure by eyeballing it, we know it has to be the side across from the 60…

00:55

…so we can use that information to solve for CE, the hypotenuse.

01:00

5 equals x square root of 3.

01:03

We want to get x alone…

01:06

…so we divide both sides by square root of 3…

01:09

…and x equals 5 over the square root of 3.

01:13

Our hypotenuse has to be 2x…so multiply that whole thing by 2 and we wind up with

01:18

10 over the square root of 3. Of course, we have this thing about leaving

01:23

square roots in the denominator… …so multiply both the top and bottom by

01:26

square root of 3 and we have 10 square root of 3 all over 3…

01:31

…which is answer B.

01:33

We didn’t even need triangle ABC. Who knows why they threw that in here.

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