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qaws [65]
3 years ago
10

Zaheer, a boy of height 1.5m was watching the entire programme. Initially he observed the top of theflagpoleat an angle of eleva

tion 300 . When he moved 10 m towards the flag post the angle of elevation of the top of the flagpole increased to 450. What is the height of the flag pole?
Mathematics
1 answer:
torisob [31]3 years ago
5 0

Answer:

h = 15.163\ meters

Step-by-step explanation:

(Assuming the correct angles are 30° and 45°)

We can use the tangent relation of the angle of elevation to find two equations, then we can use these equations to find the height of the pole.

Let's call the initial distance of the boy to the pole 'x'.

Then, with an angle of elevation of 30°, the opposite side to this angle is the height of the pole (let's call this 'h') minus the height of the boy, and the adjacent side to the angle is the distance x:

tan(30) = (h - 1.5) / x

Then, with an angle of elevation of 45°, the opposite side to this angle is still the height of the pole minus the height of the boy, and the adjacent side to the angle is the distance x minus 10:

tan(45) = (h - 1.5) / (x - 10)

So rewriting both equations using the tangents values, we have that:

0.5774 = (h - 1.5) / x

1 = (h - 1.5) / (x - 10) \rightarrow (h - 1.5) = (x - 10)

From the first equation, we have that:

x = (h - 1.5) / 0.5774

Using this value of x in the second equation, we have that:

h - 1.5 =  \frac{ (h - 1.5) }{0.5774} - 10

h + 8.5 =  \frac{ (h - 1.5) }{0.5774}

0.5774h + 4.9079 = h - 1.5

0.4226h = 6.4079

h = 15.163\ meters

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Answer:

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Step-by-step explanation:

By the SAS theorem (as you can tell by the picture) angles B and C are similar, which means they are both 47 degrees.

Since the triangles are similar (SAS theorem) then angle x must be equal to 90 degrees, because they are on a straight line both angles would need to add up to 180 degrees.

Now that you have two angles out of three, you can solve for the last angle by setting up this equation:

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nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

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