Question

A park has 2 flagpoles 20m apart. One pole is 12m high and the other is 27m high. Calculate the angle of depression from the top of the taller pole to the top of the shorter pole. Round off your answer to the nearest hundredth.

PS. Please put an illustration/drawing of the problem. Thank you!

Answer 1

θ = 37°

Explanation:

In the figure,

Distance between the poles = DC = 20 m

Height of the smaller pole = ED = 12 m

Height of the larger pole = AC = 27 m

Now, by using the properties of a rectangle,

DC = EB = 20 m

ED = BC = 12 m

Then,

AB

= AC - BC

= 27 - 12

= 15 m

We know that,

tan (θ) = opposite side / base side

Then, in △ ABE,

(Here, θ = missing angle)

tan (θ) = AB / BE

tan (θ) = 15/20

tan (θ) = 3/4

=》 θ = 37°

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Note:

Ray A & BE are parallel horizontal lines with AE as the transversal, so both the angles are equal to each other as then they will be alternate interior angles & they are represented in this question as 'θ'.

[Refer to the attachment for the figure]

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Hope it helps ⚜

Answer 2

37°

Explanation:

Here it is given that there are two flagpoles which are 20 m apart. One of them is 12m high and the other is 27m high .And we are interested in finding the angle of depression from the top of taller pole to the top of shorter pole .

For illustration, refer to the attachment or below ,

Illustration :-

`\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(0,1){6}}\put(0,0){\line(1,0){4}}\put(4,0){\line(0,1){4}}\multiput(0,4)(0.6,0){7}{\qbezier(0,0)(0,0)(0.5,0)}\qbezier(0,6)(0,6)(4,4)\put(0,6){\vector(1,0){3}}\qbezier(0.8,6)(1,5.8)(0.6,5.7)\put(1,5.6){$\theta$}\qbezier(3.2,4.42)(3,4.1)(3.25,4)\put(2.4,4.15){$\theta$}\put(-0.5,-0.5){A}\put(-0.5,6){E}\put(-0.5,3.8){D}\put(-0.8,5){15m}\put(4.3,-0.5){B}\put(-1.2,2){$ 27m $}\put(3.2,6){$ F $}\put(4.5,2){$ 12m $}\put(4.3,3.8){C}\put(-1,1.8){\vector(0,-1){1.8}}\put(-1,2.4){\vector(0,1){3.7}}\put(2,-0.5){$ 20m$}\put(2,3.5){$ 20m$}\end{picture}`

Here the angle of depression will be ,

`\longrightarrow\sf \angle_(of \ depression)=\angle FEC`

Also ,

`\sf\longrightarrow \angle FEC =\angle ECD \ (alternate \ interior \ angles)`

Let ,

`\sf\longrightarrow \angle FEC =\angle ECD = \theta \ ( say )`

Now in ∆ EDC , we have;

`\sf\longrightarrow tan\theta =(perpendicular)/(base)\\`

`\sf\longrightarrow tan\theta =(15m)/(20m)\\`

`\sf\longrightarrow tan\theta =(3)/(4)\\`

`\sf\longrightarrow \theta =tan^(-1)\bigg\lgroup (3)/(4)\bigg\rgroup\\`

`\sf\longrightarrow \underline{\boxed{\textsf{\textbf{$\boldsymbol{\theta}$= 37$^(\bf o )$ }}}}`

And we are done!