Thread: Trigonometry height and distance problems

A minar stands on a level ground.The shadow of the minar is 30 meter more when angel of elevation of sun (of sun's altitude) is than the angel of elevation of sun .Find height of the minar.

Plz solve this question.

Let the two triangles formed be ABC and ABC'. [both right angled at B]

ACB = 60 degrees, AC'B = 45 degrees.

Let AB = h, BC = x (therefore, CC' = 30)
Now in triangle ABC,

tan 60 = AB/BC
(or) tan 60 = h / x
(or) sqrt(3) = h / x
(or) x = h / sqrt(3) --------- I


In triangle ABC',

tan 45 = AB / AC'
(or) 1 = h / (AC + CC')
(or) 1 = h / (x + 30)
(or) h = x + 30
(or) h = h/sqrt(3) + 30 --------using I
(or) h = [30*sqrt(3)]/[sqrt(3) - 1]

Therefore height of the minar is (30*1.73)/(0.73) m.

Thanks for that fardeen_gen

Originally Posted by fardeen_gen

Let the two triangles formed be ABC and ABC'. [both right angled at B]

ACB = 60 degrees, AC'B = 45 degrees.

Let AB = h, BC = x (therefore, CC' = 30)
Now in triangle ABC,

tan 60 = AB/BC
(or) tan 60 = h / x
(or) sqrt(3) = h / x
(or) x = h / sqrt(3) --------- I


In triangle ABC',

tan 45 = AB / AC'
(or) 1 = h / (AC + CC')
(or) 1 = h / (x + 30)
(or) h = x + 30
(or) h = h/sqrt(3) + 30 --------using I
(or) h = [30*sqrt(3)]/[sqrt(3) - 1]

Therefore height of the minar is (30*1.73)/(0.73) m.

hey fardeen, i don't understand the last bit that i have highlighted. why is h=h/sqrt(3) + 30. I don't understand why you have to multiply the 30m by tan60 and why you have to divide that by tan35 - tan45. could you please explain why you did these calculations a.s.a.p please i'm stuck. and how is h=x+30? ty