Question

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.​

Answer 1

Answer:

In △ABD,

tan60o=BDAB

3=xh

∴   x=3h                ------- ( 1 )

Now in △ABC,

tan30o=BCAB

⇒  31=x+40h

⇒  x+40=3h

∴  3h+40=3h          [  From equation ( 1 ) ]

⇒  h+403=3h

⇒  2h=403

∴  h=203m

∴  The height of a tower is 203m.

Answer 2

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GIVEN : Tower be AB

When sun's altitude is 60°

angle ACB = 60°

Length of shadow =BC

When sun's altitude is 30°

angle ADB = 30°

Length of Shadow = DB

Shadow is 40 m when angle change from 60° to 30°

CD = 40m

angle ABC = 90°

IN TRIANGLE ABC

TAN60° = AB/CB

ROOT3 = AB/CB

CB = AB/ROOT3 ----(1)

IN TRIANGLE ABD

TAN30° = AB/DB

1/ROOT3 = AB/DB

DB = ROOT3 AB

DC + CB = ROOT3 AB

40 + CB = ROOT3 AB

CB = ROOT3 AB - 40 -----(2)

FROM (1) & (2)

AB/ROOT3 = ROOT3 AB - 40

AB = ROOT3 (ROOT3 AB) - 40 ROOT3

AB= 3AB - AB

40 ROOT3 = 3AB -AB

40 ROOT3 = 2AB

AB = 40 ROOT3 / 2

AB = 20 ROOT3

HEIGHT OF TOWER = AB = 20 ROOT3 m