Question

Question:

An observer 1.5 m tall is to 20.5 M away from a tower 22 M high determine the angle of elevation of the top of the tower from the eye of the observer.(10points)

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Answer 1

Answer :

Ꮎ = 45°

Step-by-step explanation :

Given that ;

PQ = 1.5 m

AB = 22 m

QB = PM = 20.5 m

And, PQ = MB = 1.5 m

AM = AB - BM

     = ( 22 - 1.5 ) m

    = 20.5 m

Now,

tan Ꮎ = $\bf \frac{AM}{PM}$

tan Ꮎ = $\sf \frac{20.5}{20.5}$

tan Ꮎ = 1

Hence,

The angle of elevation,i.e., Ꮎ =  45°

Answer 2

Answer

 

Refer the attachment for figure.

Step-by-step explanation :

Let AB be the observer and CD be the tower.

Now,  AB = 1.5 m and CD = 22 m

BD = 20.5 m

Draw AL perpendicular to CD. Then,

AL = BD = 20.5 m

LD = AB = 1.5 m

( Opposite sides of parallelogram are equal )

Now,

CL = CD - LD = 22 - 1.5

= 20.5

Let Angle CAL be theta.

In ∆ CLD,

$\tt{tan\theta=\frac{CL}{AL}}$

$\tt{=\frac{20.5}{20.5}= 1}$

tan theta = tan 45°



Hence,

Theta =45 °

Angle of Elevation is 45°.





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