A flagpole stands on a building of height 450 ft and an observer on a level ground is 300 ft from the base of the building. The angle of elevation of the bottom of the flagpole is 30° and the height of the flagpole is 50 ft. If θ is the angle of elevation of the top of the flagpole, then tan θ is equal to

Question

A flagpole stands on a building of height 450 ft and an observer on a level ground is 300 ft from the base of the building. The angle of elevation of the bottom of the flagpole is 30° and the height of the flagpole is 50 ft. If θ is the angle of elevation of the top of the flagpole, then tan θ is equal to (A)  (4/3√3)  (B)  (√3/2)  (C)  (9/2)  (D)  (√3/5)

Tardigrade Admin · Accepted Answer

In Δ DCE, tan 30° =(150/CD)⇒ CD=(150/1/√3) ⇒ CD=√3× 150 Now, in Δ DCF, tan θ =(DF/CD)=(200/√3.150)=(4/3√3)

$\frac{4}{3 3}$

$\frac{3}{2}$

$\frac{9}{2}$

$\frac{3}{5}$

$\frac{4 3 + 1}{6}$

In $Δ D CE,$
$tan 30^{\circ} = \frac{150}{C D} \Rightarrow C D = \frac{150}{1/ 3}$
$\Rightarrow$ $C D = 3 \times 150$
Now, in $Δ D CF,$ $tan θ = \frac{D F}{C D} = \frac{200}{3 .150} = \frac{4}{3 3}$

33​4​

23​​

29​

53​​

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