Question

A vertical flagpole, AB,has a height of 42m. The points B, C and D lie on level ground, and BCD is a straight line Point C is 24 m from the base of the flagpole. The angle of elevation of the top of the flagpole from point D is 48 degrees
How far apart are points C and D?

Answer 1

Answer: 13.8

Step-by-step explanation:

42/tan(48˚)=37.81696986

37.81696986-24=13.81696986

=13.8

Answer 2

Answer:

Points C and D are 13.82 meters apart.

Step-by-step explanation:

Given : Flag AB, with point B , C ,D lying on ground.

BD = 24 m . AB = 42 m

To find; CD

Solution:

In triangle ABD, with angle of elevation from D to top of flag pole is 48 degrees.

$\tan \theta =\frac{perpendicular}{base}$

$\tan 48^o=\frac{AB}{BD}$

$1.1106=\frac{42 m}{BD}$

$BD=\frac{42 m}{1.1106}=37.82 m$

From the figure:

BD = BC + CD

37.82 m = 24 m + CD

CD = 37.82 m - 24 m = 13.82 m

Points C and D are 13.82 meters apart.