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3 years ago

6

A cell tower has the hieght of 70 meters. The Tower casts a shadow of 45 meters on the ground. What is the angle of elevation of

the shadow to the top of the cell tower?

1 answer:

laiz [17]3 years ago

4 0

Answer:

$\theta=0.99^{\circ}$

Step-by-step explanation:

Given that,

The height of a cell tower, P = 70 m

The tower casts a shadow of 45 meters on the ground.

We need to find the angle of elevation of the shadow to the top of the cell tower. We can use trigonometry to find it.

$\tan\theta=\dfrac{P}{B}$

Put all the values,

$\tan\theta=\dfrac{70}{45}\\\\\theta=\tan^{-1}\left(\frac{70}{45}\right)\\\\\theta=0.99^{\circ}$

Hence, the angle of elevation is equal to $0.99^{\circ}$ .

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Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

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Step-by-step explanation:

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At n=1,

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At n=2,

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At n=3,

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The second explicit equation is

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At n=1,

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At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

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At n=2,

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At n=3,

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At n=2,

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At n=3,

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