A tower(t) h meters high stands on the top of a mound(m) At a point on the ground m subtends<a and t <b.prove

Question

that
( h tan a)/{tan(a+b)-tan a}=height of the mound(m)

Answer 1

I have to climb up the tower to answer your question!

Answer 2

Let A be the point on the ground vertically below the tower and B be the point where tower and mound subtend angles Let top of the mound be C and that of the tower be D.
From triangle ABC,
tan a = AC/AB = H/AB (where H is the height of the mound)
or, AB = H/tan a (1)
From triangle ADC,
tan (a+b) = AD/AB = (H+h)/AB
or AB = (H+h)/tan (a+b) (2)
From (1) & (2) above,
H/tan a = (H+h)/ tan (a+b)
or, tan (a+b) / tan a = (H+h)/H = 1+h/H
or h/H = tan (a+b)/tan a - 1 = [tan (a+b) - tan a] / tan a.
or h tan a /[tan (a+b) - tan a] = H

Answer 3

Boy your question is rather mixedup.

the answer & solutions provided would be correct only if the total angle subtended by t(with ground)=(a+b). Else it does not seem to be possible.

Try it.

Answer 4

let the height of the moundbe x
et thehorizontal dist. bet. theobserver and the mound bey
the equations are
x/y=tan a
x=ytana........(1)

h+x/y=tan(a+b)
h+x=y tan (a+b).....(2)
dividing (1) by (2)
x/h+x=ytana/ytan(a+b)
x tan (a+b)=(h+x) tan a
x tan(a+b)-xtana=h tan a
x=h tan a/[tan(a+b)-tan a]
hence the question

Answer 5

m+h = d tan(a+b) , d is the distance from observer to mound.
m = d tan a

From the above two equations, eliminate d to obtain an expression for m in terms of a,b,&h using algebra and trig.

From eq.(2) d = m/tan(a). Now, plug this into the first eq. to get

m+h = [m/tan(a)] tan(a+b), so
m tan(a) + h tan(a) = m tan(a+b)
m[tan(a)-tan(a+b)] = - h tan(a)

Finally, m = -h tan(a)/[tan(a)-tan(a+b)], or

m = h tan(a) / [tan(a+b)-tan(a)] as you wished to show.