You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
2009 DAT QRC
Started by cobbby
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.
Attachments
Last edited:
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.
Advertisement - Members don't see this ad
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 DATI 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.
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.
Lnguyen - would you say its important to memorize the unit circle for the DAT?
Its on the bootcamp QR formula sheet, and yes you should know itLnguyen - 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.Lnguyen - would you say its important to memorize the unit circle for the DAT?
Similar threads
- Replies
- 0
- Views
- 871