Trig question (1 Viewer)

[acmilan]

Here is a random trig question i found in this science magazine that I think is a little different from normal. (Its not necessarily hard) Im not sure if this identity is taught, and is needed for one method of getting the answer so ill post it:

tan(90 - x) = 1/tan(x) (or tan(pi/2 - x) = 1/tan(x) in radians)

Heres the question:

Someone wants to calculate the height of a rugby post. He has a device for measuring distance along the ground and another for measuring sighted angles. He starts at the base of the post, measures 10 standard units in a straight line along the ground, measures the angle from the ground to the top of the post and records the data. Obviously this is sufficient to calculate the height but John decides to take two more sets of measurements. He walks a further 10 units in the same direction and measures the angle. He then walks another 10 units and measures the angle again. What is the height of the pillar (in standard units) given that the sum of the three measured angles is 90 degrees?

 

[shafqat]

tan(90 - x) = cotx = 1/tanx. I think complementary angles are taught.

 

[who_loves_maths]

hi acmilan,

nice question. although, i didn't see where you actually needed to use tan(90 - x) = 1/tan(x); but it's probably cause i rushed it.

but anyhow, here's my solution (i figured it'd take too long to type up the solution, so i just wrote it down on paper):

Applied Trigonometry

there might be easier way(s) of doing it out there; but i think this method is pretty straightforward anyway.

P.S. for the trigonometric identity there in my solution; it is actually "taught" at the HSC level - and like i illustrated in my solution, isn't very hard to derive. however, just for reference to that particular identity i used, go to: GREEN Fitzpatrick (3 Unit), Exercise 27(d), Question 26, pg. 174 ...


Edit: plz let me know if i've made any mistakes in my solution.

 

[acmilan]

Nice work you got it right

There are 2 other ways ive seen it done. Ill post other solutions tomorrow if no one else posts them first.

 

[KFunk]

How not to do the question:

sina = h/AH --> AH= h/sina (1)

AH/sinb = BH/sin(90-a) --> AH = (sinb.BH)/cosa (2)

Equating (1) and (2)

h = tana.sinb.BH (3)

considering the Δ BHF, cosb= 20/BH hence BH= 20/cosb (4)

Subing (4) into (3)

h = 20.tana.tanb , but tana = h/10 and tanb= h/20 so

h = h²/10 , h² = 10h ∴ h = 10

 

[acmilan]

Something funky is goin' on ... fluke?

From the looks of it I suspect it may be.

 

[who_loves_maths]

Originally Posted by KFunk
If they phrased the question such that it implied the use of all three angles, would you have to use such a method? It's a lot easier if you stop before you go past two:

Let the first two angles be a and b at the points A and B which are 10 and 20 metres from the foot of the post F, respectively. Name the top of the post H. It then follows that

sina = h/AH --> AH= h/sina (1)

AH/sinb = BH/sin(90-a) --> AH = (sinb.BH)/cosa (2)

Equating (1) and (2)

h = tana.sinb.BH (3)

considering the Δ BHF, cosb= 20/BH hence BH= 20/cosb (4)

Subing (4) into (3)

h = 20.tana.tanb , but tana = h/10 and tanb= h/20 so

h = h2/10 , h2 = 10h ∴ h = 10

You can extend the same idea to the third angle but it's really just a longer way of doing the same thing. So as I was asking, would the above be a valid method given the phrasing of the question?

EDIT: Something is wierding me out in what went on above. You could measure a distance of 10 and then 20 meters from a post of any height. Something funky is goin' on ... fluke?

Originally Posted by acmilan
From the looks of it I suspect it may be.

sorry, guys, not fluke... just flaw.

AH/Sin(b) does NOT equal BH/Sin(90 -a) = BH/Cos(a) ;

instead, i believe AH/Sin(b) = BH/Sin(180 -a) = BH/Sin(a) ?

 

[KFunk]

haha, well at least my mind is put at ease. I thought that being flu-ridden wouldn't stop me attacking a maths question, I thought wrong .

 

[acmilan]

Ill give one other method now. Its not the most elegant of the two I have but it still works:

Let the respective angles be a, b and c. Let h be the height and u be the 'standard unit'

tan a = h/(10u)

tan b = h/(20u)

tan c = h/(30u)

Let x = h/u

tan a = x/10 -> a = arctan(x/10)

tan b = x/20 -> b = arctan(x/20)

tan c = x/30 -> c = arctan(x/30)

Also a + b + c = pi/2

So arctan(x/10) + arctan(x/20) + arctan(x/30) = pi/2

Let f(x) = arctan(x/10) + arctan(x/20) + arctan(x/30)

So we need to find a value of x for which f(x) = pi/2. A known property is that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2, hence we can take x = 10

Hence x = h/u = 10 thus h = 10u or 10 standard units.

Like I said, not the best method as you'd have to know the property that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2 to get to the solution simply

 

[KFunk]

Another go :

Let the angles be a, b and c at the points A, B and C which are 10, 20 and 30 metres from the foot of the post F, respectively. Name the top of the post H.

Since a+b+c=90 --> cos(a + b + c) = 0

cos(a+b)cosc - sin(a+b)sinc = 0

cosa.cosb.cosc -sina.sinb.cosc - sina.cosb.sinc - cosa.sinb.sinc = 0

where cosa = 10/AH , cosb = 20/BH , cosc = 30/CH, sina = h/AH, sinb = h/BH, sinc = h/CH

Then multiply through by AH.BH.CH since all terms have a denominator of AH.BH.CH

6000 - 30h² - 20h² - 10h² = 0

60h² = 6000

h² = 100, h = 10

 

[who_loves_maths]

Originally Posted by acmilan
Ill give one other method now. Its not the most elegant of the two I have but it still works:

Let the respective angles be a, b and c. Let h be the height and u be the 'standard unit'

tan a = h/(10u)

tan b = h/(20u)

tan c = h/(30u)

Let x = h/u

tan a = x/10 -> a = arctan(x/10)

tan b = x/20 -> b = arctan(x/20)

tan c = x/30 -> c = arctan(x/30)

Also a + b + c = pi/2

So arctan(x/10) + arctan(x/20) + arctan(x/30) = pi/2

Let f(x) = arctan(x/10) + arctan(x/20) + arctan(x/30)

So we need to find a value of x for which f(x) = pi/2. A known property is that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2, hence we can take x = 10

Hence x = h/u = 10 thus h = 10u or 10 standard units.

Like I said, not the best method as you'd have to know the property that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2 to get to the solution simply

lol, i have to say that this method is almost identical to the one i posted... except this method doesn't give a reason as to why one would gratuitously be expected to know that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2 .

i am confident that this result is/can be directly derived from the identity i found in my solution anyways, so at least that provides a reason for it i suppose.

in addition, i don't understand why this solution specifies that x=h/u ; 'u' is just the symbol for a unit, it's not a pronumeral and does not have a value associated with it... generally, in all geometry problem, no unit is specified anyhow, only numbers, meaning that the final number you get as an answer is just that number of units - and it's up to the participant to convert it; so i think the 'u' is pretty much pointless here.

 

[acmilan]

lol, i have to say that this method is almost identical to the one i posted... except this method doesn't give a reason as to why one would gratuitously be expected to know that arctan(1) + arctan(1/2) + arctan(1/3) = pi/2 .

i am confident that this result is/can be directly derived from the identity i found in my solution anyways, so at least that provides a reason for it i suppose.

in addition, i don't understand why this solution specifies that x=h/u ; 'u' is just the symbol for a unit, it's not a pronumeral and does not have a value associated with it... generally, in all geometry problem, no unit is specified anyhow, only numbers, meaning that the final number you get as an answer is just that number of units - and it's up to the participant to convert it; so i think the 'u' is pretty much pointless here.

Yeah i agree its not the best method and probably not the process i wouldve went through (or thought of) to get the answer.

 

[who_loves_maths]

acmilan & co. ,

okay well i've tried to look for that elusive 'more elegant' solution of yours, and so far i have come up with another solution of this problem, but am not sure whether it's the more elegant one. however, this new solution is a much quicker method than the other one i had before.
i can't be bothered wasting more time on this problem so i'll just be satisfied with two solutions to-date thus far;

here is my new one:

No Trig. !
or,
With a tad of Trig.

{i hope you guys will like this one, seeing that the method is almost 'separate' from any geometric techniques... provides another perspective and it ties in a 4-unit topic as well } (excuse my enthusiasm)

 

[acmilan]

haha you continue to impress me who_loves_maths.

The other solution given is pretty similar to KFunk's except using tan:

a + b + c = pi/2

a + b = pi/2 - c

tan(a + b) = tan(pi/2 - c)

tan(a + b) = 1/tan(c)

(tan(a)+ tan(b))/(1 - tan(a)tan(b)) = 1/tan(c)

(h/10 + h/20)/(1 - (h/10)(h/20)) = 30/h

This ends up giving 1200h² = 120000 and hence h = 10

 

[KFunk]

{i hope you guys will like this one, seeing that the method is almost 'separate' from any geometric techniques... provides another perspective and it ties in a 4-unit topic as well } (excuse my enthusiasm)

Nice application, it's always a good exercise to try and find more ways of doing a question. Also, your enthusiasm is understood, it's quite a cool method . One trick I've noticed you using which I never knew you could use is the following:

If p/q = ∞ , then q=0

Is that a normal conclusion that you can always use or are there some provisos that you need to be weary of?

 

[who_loves_maths]

Originally Posted by KFunk
If p/q = ∞ , then q=0

Is that a normal conclusion that you can always use or are there some provisos that you need to be weary of?

if you have p(x)/q(x) -> infinity, and let's say both functions in 'x' are polynomials, then you have to first reduce/simplify the quotient of p(x)/q(x) to its lowest degree (ie. cancels out all common factors, etc).... and then you can apply the q(x) = 0 test/rule.
i'm certain that this is true for all polynomial 'p' and 'q' {which is the case for the problem of this particular thread};
and i'm pretty sure that this is also true for the other types of functions that we encounter in HSC mathematics... however, beyond that level, i can't say i know enough to comment in terms of q(x) = 0 being the ONLY solution(s).

for example, intuitively, you can pose the question in reverse: if p(x) and q(x) are any functions in 'x', and some values {a(0), a(1), a(2), ...} are zeros of q(x) which are unique to q(x) {ie. NOT zeros of p(x)}; then for x = {a(0), a(1), a(2), ...}, p(x)/q(x) clearly approaches infinity.
and so reversing this again: If given that p(x)/q(x) -> infinity, find 'x': -----> clearly x = {a(0), a(1), a(2), ...} is a solution subset.

Edit: from the perspective of Graphing, you can say that the behaviour of F(x) = p(x)/q(x) is asymptotic at x = {a(0), a(1), a(2), ...}

 

[who_loves_maths]

Originally Posted by acmilan
haha you continue to impress me who_loves_maths.

The other solution given is pretty similar to KFunk's except using tan:

a + b + c = pi/2

a + b = pi/2 - c

tan(a + b) = tan(pi/2 - c)

tan(a + b) = 1/tan(c)

(tan(a)+ tan(b))/(1 - tan(a)tan(b)) = 1/tan(c)

(h/10 + h/20)/(1 - (h/10)(h/20)) = 30/h

This ends up giving 1200h2 = 120000 and hence h = 10

wow, that's very neat and natty acmilan

darn, wish i'd have thought of that. it's the equivalent of using the trig. identity i used, except mine was too general and this method is so much faster, cleverer, and more expedient.

nice!