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Question
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

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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?

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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.

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