Question

From the top of a building of 20m high the angle of elevation of the top of a monument is 45⁰ and the angle of depression of the foot of the monument is 30⁰. Find the height of the monument in nearest metre.​

Answer 1

Answer:

Given,

BC=DE=20mtan75=BCBD=20BD=3.7321⇒BD=74.642tan(∠ADB)=BDAB⇒1=74.642AB

Height of tower =AB+BC

=94.642 m 

Hence, this is the answer.

Answer 2

Answer:

let AB be the building (with A at top) and CD be the monument (with C at top)

let AE║BD, AE = BD and AB = ED

In ΔAEC,

           tan 45° = CE/ AE

        ⇒ CE = AE             (tan 45° = 1) ................................ 1

alternate interior angles are equal

          ⇒ EAD = BDA = 45°

In ΔABD

          tan 45° = AB/ BD 

                 ⇒ AB = BD          (tan 45° = 1) ................................ 2

Since AE = BD

              AB = CE                 (from 1 and 2)

thus the height of monument = CE + ED = CD = 20 + 20

                                                    = 40