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lesya692 [45]
3 years ago
14

A school flag poll casts a 16 foot shadow on the lawn. A teacher stood at the shadows edge and measure the angle of elevation to

the top of the poll at 62°. How tall is the poll?

Mathematics
1 answer:
vovikov84 [41]3 years ago
6 0

Answer:

30.1 feet.  

Step-by-step explanation:

We have been given that a school flag poll casts a 16 foot shadow on the lawn. A teacher stood at the shadows edge and measure the angle of elevation to the top of the poll at 62°. We are asked to find the height of the tall.

First of all, we will draw a diagram to represent our give scenario as shown in the attachment.

We can see that side length with 16 feet is adjacent side and height of the pole (h) is opposite side to angle of elevation.

We know that tangent relates opposite side of a right triangle to its adjacent side.

\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}

\text{tan}(62^{\circ})=\frac{h}{16}

h=16\cdot \text{tan}(62^{\circ})

h=16\cdot 1.880726465346

h=30.0916234\approx 30.1

Therefore, the poll is approximately 30.1 feet tall.

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ehidna [41]

<u>Given:</u>

The coin wrapper volume = 27,480 mm³.

The coin diameter = 21.18 mm.

There are 50 such coins.

<u>To find:</u>

The thickness (height) of 1 coin.

<u>Solution:</u>

The 50 coins form a cylinder with a volume of 27,480 mm³.

If the diameter is 21.18 mm, the radius of a coin will be \frac{21.18}{2} = 10.59 mm.

The volume of a cylinder = \pi r^{2} h.

The volume of all 50 coins is given as 27,480 mm³. We need to determine the height of a single coin.

50( \pi r^{2} h)=27,480.

50(3.1415)(10.59^{2} )(h) = 27,480.

(17,615.662)(h)=27,480.

h= \frac{27,480}{17,615.662} = 1.5599 mm.

So the thickness of 1 coin is 1.5599 mm.

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