Thread: Trigonometry height and distance problems

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