2009 DAT QRC

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I don't have the exam on me, but that exam is riddled with errors - just a heads up (Not saying those questions have errors, but if you think something is inccorectly marked - it may be).
 
Hello
Anybody has the explanation for 2009 DAT QR SECTION specially # 4,6,14?
Thanks


4.
tan(2θ) = AB/DB
AB = DB*tan(2θ)
AB = DB*2tanθ/(1-tan^2(θ))
AB = DB*2(DE/FE)/(1-(DE/FE)^2)

Right now we have to find DE, FE and DB
x^2 + y^2 = 25
x^2 = 25 - y^2 = 25 - 3^2 = 16
x = ±4
So point D is (4, 3). Therefore FE is 4, DE is 3 and DB is 12 - 4 = 8

AB = 8*2(3/4)/(1-(3/4)^2) = 192/7
AC = AB + BC = 192/7 + 3 = 30 3/7

AC starts from y = 0 and goes up 30 3/7 so the y is 30 3/7

6.
The goal is to find the line equation for DE.

Slope of AB = -3/2
Thus slope of BC, CD and DE are 3/2, -3/2 and 3/2 respectively. This is due to the nature of incident and refraction angles.

BC: y = 3/2x + b
b= 0 - 3/2(2) = -3
y = 3/2x -3

Point C coordinate is (x, 10)
Finding x: 10 = 3/2x -3 , x = 26/3
C (26/3, 10)

CD: y = -3/2x + b
b = 10 + 3/2(26/3) = 23
y = -3/2x + 23

Point D coordinate is (10, y)
Finding y: y = -3/2(10) + 23, y = 8
D (10, 8)

Note: it's not possible for CD to hit the bottom of the box.

DE: y = 3/2x + b
b = 8 - 3/2(10) = -7
y = 3/2x - 7
Point E coordinate is (x, 0)
Finding x: 0 = 3/2x - 7, x = 14/3

14.
You use reference angle here
OP = SQRT(8^2+15^2) = 17
Cos(α) = - cos(γ) = -(8/17)

Cosine of any angle in second quadrant is negative.
 

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Dayum, dude went hero mode. Super generous post

4.
tan(2θ) = AB/DB
AB = DB*tan(2θ)
AB = DB*2tanθ/(1-tan^2(θ))
AB = DB*2(DE/FE)/(1-(DE/FE)^2)

Right now we have to find DE, FE and DB
x^2 + y^2 = 25
x^2 = 25 - y^2 = 25 - 3^2 = 16
x = ±4
So point D is (4, 3). Therefore FE is 4, DE is 3 and DB is 12 - 4 = 8

AB = 8*2(3/4)/(1-(3/4)^2) = 192/7
AC = AB + BC = 192/7 + 3 = 30 3/7

AC starts from y = 0 and goes up 30 3/7 so the y is 30 3/7

6.
The goal is to find the line equation for DE.

Slope of AB = -3/2
Thus slope of BC, CD and DE are 3/2, -3/2 and 3/2 respectively. This is due to the nature of incident and refraction angles.

BC: y = 3/2x + b
b= 0 - 3/2(2) = -3
y = 3/2x -3

Point C coordinate is (x, 10)
Finding x: 10 = 3/2x -3 , x = 26/3
C (26/3, 10)

CD: y = -3/2x + b
b = 10 + 3/2(26/3) = 23
y = -3/2x + 23

Point D coordinate is (10, y)
Finding y: y = -3/2(10) + 23, y = 8
D (10, 8)

Note: it's not possible for CD to hit the bottom of the box.

DE: y = 3/2x + b
b = 8 - 3/2(10) = -7
y = 3/2x - 7
Point E coordinate is (x, 0)
Finding x: 0 = 3/2x - 7, x = 14/3

14.
You use reference angle here
OP = SQRT(8^2+15^2) = 17
Cos(α) = - cos(γ) = -(8/17)

Cosine of any angle in second quadrant is negative.
 
I wouldn't bother with this problem. There is no way you can finish them in 1 minute. Furthermore, every problem is weigh the same. Focus on the fundamentals and concepts that are taught in Math Destroyer. Don't focus on something that is almost impossible to do in 1 minute.
 
I wouldn't bother with this problem. There is no way you can finish them in 1 minute. Furthermore, every problem is weigh the same. Focus on the fundamentals and concepts that are taught in Math Destroyer. Don't focus on something that is almost impossible to do in 1 minute.
Ya, its pretty in depth. I would still understand it, but its unlikely to see something of this magnitude on the real DAT
 
Lnguyen - would you say its important to memorize the unit circle for the DAT?
Yes you should. It's very helpful. I'd memorize the first quadrant and calculate the rest using reference angle instead of memorizing the whole circle.
 
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