Question

In the figure, given below, it is given that AB is perpandiculer to BD and is of length X metres. DC = 30 m, ∠ADB = 30° and ∠ACB = 45°. Without using tables, find X.

Answer 1

In ΔABC, we have

`tan 45^circ = (AB)/(CB) = X/(CB)`  ...`[∵ tan theta = "Perpendicular"/"Base"]`

`1 = X/(CB)`   

`\implies` CB = X   ...(i)

Again in ΔADB, we have

`tan 30^circ = (AB)/(DB) = X/(DB)`

`\implies 1/sqrt(3) = X/(DB)`

∴ `DB = sqrt(3) xx X`  ...(ii)

But DB = DC + CB

∴ `sqrt(3)X = 30 + X`

`\implies sqrt(3)X - X = 30`

`\implies X(sqrt(3) - 1) = 30`

`\implies X = 30/(sqrt(3) - 1)m`

∴ `X = (30(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`

= `(30(sqrt(3) + 1))/(3 - 1)m`

= `(30(sqrt(3) + 1))/2`

= 15 (1.732 + 1) m.

= 15 × 2.732

= 40.98 m.