Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 5 metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively 30° and 60°. Find the height of the tower.

Answer 1

Answer:

The height of the tower is 2.5 m  

Step-by-step explanation:

Given :  

Height of flagstaff, AB = 5 m  

Angle of elevation from the top of the flagstaff (θ), ∠ADC = 60°

Angle of elevation from the bottom of the flagstaff ,(θ), ∠BDC = 30°

Height of tower ,BC = h m

Let CD = x m

In right angle triangle, ∆BCD ,

tan θ  = P/ B

tan 30° = BC/CD

1/√3 = h/x

x = √3h ………..(1)

In right angle triangle, ∆ACD,

tan θ  = P/ B

tan 60° = AC/CD

√3 = (AB + BC)/CD

√3 = (5 + h)/x

√3x = (h + 5)

√3 × √3h = (h + 5)

[From eq 1]

3h = (h + 5)

3h - h = 5  

2h = 5  

h = 5/2

h = 2.5 m

Hence, the height of the tower is 2.5 m  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION}}}$

Let AD be the Flagstaff, 5m high & DC be the tower. B be the point on the plane from where the angles of elevation of the top(A) and bottom of the Flagstaff (D) are 60° & 30° respectively. Join C& B. Then we get two ∆ABC & ∆BCD with right angle at C. We are to find the height of the tower DC.

Given,

AD= 5m

∠ABC= 60°&

∠DBC= 30°

In ∆ABC,

$tan∠</strong><strong>A</strong><strong>B</strong><strong>C</strong><strong> = tan60 \degree = \frac{</strong><strong>AB</strong><strong>}{BC} \\ \\ = > \frac{ad + DC}{BC} = \frac{5 + DC}{BC} \: or \: \\ = > </strong><strong>BC</strong><strong> = \frac{5 + </strong><strong>DC</strong><strong>}{ \sqrt{3} }$

In ∆BCD,

$tan∠</strong><strong>D</strong><strong>B</strong><strong>C</strong><strong> = tan30 \degree = \frac{</strong><strong>DC</strong><strong>}{</strong><strong>BC</strong><strong>} \\ \\ or \\ = > </strong><strong>BC</strong><strong> = \frac{</strong><strong>DC</strong><strong>}{tan30 \degree} = </strong><strong>DC</strong><strong> \sqrt{3}$

Equating the values of BC,

$DC \sqrt{3} = \frac{5 + DC}{ \sqrt{3} } \\ \\ = > 3DC= 5 + DC \\ \\ = > 3dc - DC = 5 \\ \\ = > 2DC = 5 \\ \\ = > DC = \frac{5}{2} = 2.5 \\ \\ = > DC = 2.5m$

Hence, the height of the tower is 2.5m.

Hope it helps ☺️