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abruzzese [7]
3 years ago
14

Does anyone know how to do 15 and 16?????

Mathematics
2 answers:
Julli [10]3 years ago
3 0
This is how you answer number 15 and 16

Ber [7]3 years ago
3 0
ANSWERS- 15.) the second part would be 46.25 i don’t know the first part so sorry though

16.) x=33.3 and 3x+10=57.3


Idk if any of these are right though I haven’t done this in a long time sorry if I made mistakes.
STEPS-
You would find out the measure of the type of line which would be 180 since it’s straight. Then you would solve it algebraically so you would do 4x-5=180, add 5 to both sides (5 would cancel out & 180 would be 185) the divide 185 by 4x and you would get x = 46.25. Sorry I don’t know the other equation for 15.

For 16.) you would turn the equation into 3x-10=90 since it’s a 90 degree angle. Add 10 to both sides so that they cancel each other out then add 10 to 90 and then you should get 100. Now your equation should look like this 3x=100. Next do 100➗3x= 33.3. Now your answer is x=33.3 and the second part would be 90-33.3= 57.3.






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One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on
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a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

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d. Height of antenna  =  237.08 ft

Step-by-step explanation:

​The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°  

distance from her position to the top of the building  = 2475 ft

distance from her position to the top of the flag  = 2615 ft

​(a) How far away from the base of the building is the surveyor​ located?​

using the SOHCAHTOA principle

cos 34° = c/2475

c =  0.8290375726  × 2475

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(b) How tall is the​ building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b =  0.5591929035  × 2475

b =  1384.0024361

b =  1384.00 ft

​(c) What is the angle of elevation from the surveyor to the top of the​ antenna?

let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ =  0.784665392

∅ = cos-1  0.784665392

∅ =   38.310258303

∅ =  38.31°

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height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) =  0.6199159917  × 2615

(a + b) =  1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00  =  237.08031822 ft

Height of antenna  =  237.08 ft

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