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This a Class X based question.
A two storeyed building has the height of the lower storey 12 metre and that of upper storey 13 metre. Find at what distance the two storeys subtend equal angles to an observer's eye at height 5 metre from the ground."

Please do not reject it as a home work question for it is a sum in another book (I practise) and not the school given one.

First of all I'm not able to draw an appropriate diagram and solving it is altogether a different job!
I'll wait for a positive reply."

This indeed is an interesting question.  It looks like, spelled properly in American, as:

"A two storied building has the height of the lower story 12 meters and that of upper story 13 meters. Find at what distance the two stories subtend equal angles to an observer's eye at height 5 meters from the ground."

If I read I correctly, it says that the height of the first floor is 12 meters and the height of the second floor is 13 meters.  That makes the total height 12+13=25 meters.

The height if the far leg of a triangle and the distance away is the near leg.
To make right triangles, draw a horizontal line from the observer's eye to the building.
Label this as x.

I will use tan() to be the cotangent (the near side over the far side on a right triangle).
I will use arctan() to be the inverse of the tan().  I will use A for the first angle and
B for the second angle.

The first story starts at 5 meters below and goes to 12-5 = 7 meters above.  That gives two right triangles with x as the horizontal near side to the angle, so call them x1 and x2.
It is known that tan(x1) = 5/x and the tan(x2) = 7/x.  Thus, the total angle A would be
A = arctan(5/x) + arctan(7/x).

The second story starts at 7 meters above the line of eyesight and goes to 13+7=20 meters above.  This also generates two right triangles, known as y1 and y2.  The distance is the difference between them.  The angle y1 is given by the height of the bottom, which would be
cnt(y1) = 7/x.  The other angle would be given by the one drawn to the top.   It would be
tan(y2) = 20/x.  Thus, the second angle B would be given by B = arctan(20/x) - arctan(7/x).

We want angle A to be the same as angle B.  This means we need
arctan(5/x) + arctan(7/x) = arctan(20/x) - arctan(7/x).

Subtracting arctan(5/x) + arctan(7/x) from both sides gives
0 =  arctan(20/x) - arctan(7/x) - arctan(5/x) - arctan(7/x).

Combining the arctan(7/x) terms gives 0 =  arctan(20/x) - 2*arctan(7/x) - arctan(5/x).

Putting in Excel and solving using the Secant Method* with x1 as 40 and x2 as 50 gives
40   -0.007198718
50   0.002645842
47.31238204   0.000875665
45.98288075   -0.000188547
46.21842913   1.03282E-05
46.20619635   1.13748E-07
46.20606013   -6.9613E-11
46.20606021   4.71845E-16
46.20606021   0

That says that x needs to be really close to 46.2 feet,
which, using 12*.206 for inches is 46' 2.47", which is roughly  46' [2 1/2]".
That forty six feet and two and a half inches.

* The secant method says that x3 = x1 - y1(x2-x1)/(y2-y1) where (x1,y1) is one point,
(x2, y2) is another point, and (x3,y3) is the solution.  It assumes that y3 is suppose
to be 0, which is what the problem is getting to after doing it for several times.
Each time it is done again, (x1,y1) is replaced with (x2,y2) and then (x2,y2) is
replaced with (x3,y3).

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Scott A Wilson


I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?


Experience in the area; I have tutored people in the above areas of mathematics for over two years in I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.

My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.

Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.

Past/Present Clients
My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more.

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