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Alex_Xolod [135]
3 years ago
13

A person can see the top of a building at an angle of 65°. The person is standing 50 ft away from the building and has an eye le

vel of 5 ft. How tall is the building to the nearest tenth of a foot?

Mathematics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

112.2ft

Step-by-step explanation:

As can be seen in the attached diagram, The top of the building is at C.

The distance from the man to the building is |BD| and the height of the man is |AB| where his eye level is at B.

We are required to find the height |CE| of the building.

Now |AB|=|DE|, (opposite sides of a rectangle).

Tan 65° = |CD|/59

|CD|=59 XTan65°=107.23

The height of the building |CE|=|CD|+|DE|=107.23+5=112.23ft

The height=112.2ft ( to the nearest tenth)

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Answer:

B

Step-by-step explanation:

Given that r is inversely proportional to s then the equation relating them is

r = \frac{k}{s} ← k is the constant of proportion

To find k use the condition r = 16 when s = 3

k = rs = 16 × 3 = 48

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BRAINLIEST. NEED HELP NOW!!!<br> What is the value of x?
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Answer:

x=10

Step-by-step explanation:

these are similar triangles

8             12

------   = -----------

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16x-12x +8 = 12x-12x+48

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x =10

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3 years ago
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Answer:

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Step-by-step explanation:

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Hello there.

<span>(a) A color printer prints 11 pages in 7 minutes. How many pages does it print per minute? 
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