from an observer o, the angles of elevation of the bottom and the top of a flagpole are 40° and 45° respectively.find the height
1 answer:
Answer:
Take a look of the image below, we can think on this problem as a problem of two triangle rectangles.
We can see that both triangles share the adjacent cathetus, then the height of the flagpole is just the difference between the opposite cathetus.
Remember the relation:
Tan(a) = (opposite cathetus)/(adjacent cathetus)
So, if we define H as the height of the cliff
X as the distance between the observer and the cliff
and h as the height of the flagopole
we can write:
tan(40°) = H/X
tan(45°) = (H + h)/X
Notice that we have two equations and 3 variables (we should have the same number of equations than variables) then here is missing information, and we can't get an exact solution for the height of the flagpole.
But we can write it in terms of the height of the cliff H, or in terms of the distance between the observer and the cliff.
We want to find the value of h.
If we take the quotient between both equations, we get:
Tan(45°)/Tan(40°) = (H + h)/H
1.192 = (H + h)/H
1.192*H = H + h
1.192*H - H = h
0.192*H = h
So the height of the flagpole is 0.192 times the height of the cliff.
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Answer:
(a) The probability that X is at most 30 is 0.9726.
(b) The probability that X is less than 30 is 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is 0.7406.
Step-by-step explanation:
We are given that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked. A random sample of 200 shafts is taken.
Let X = <u><em>the number among these that are nonconforming and can be reworked</em></u>
The above situation can be represented through binomial distribution such that X ~ Binom(n = 200, p = 0.11).
Here the probability of success is 11% that this much % of all steel shafts produced by a certain process are nonconforming but can be reworked.
Now, here to calculate the probability we will use normal approximation because the sample size if very large(i.e. greater than 30).
So, the new mean of X, =
=
= 22
and the new standard deviation of X, =
=
= 4.42
So, X ~ Normal()
(a) The probability that X is at most 30 is given by = P(X < 30.5) {using continuity correction}
P(X < 30.5) = P( <
) = P(Z < 1.92) = <u>0.9726</u>
The above probability is calculated by looking at the value of x = 1.92 in the z table which has an area of 0.9726.
(b) The probability that X is less than 30 is given by = P(X 29.5) {using continuity correction}
P(X 29.5) = P(
) = P(Z
1.70) = <u>0.9554</u>
The above probability is calculated by looking at the value of x = 1.70 in the z table which has an area of 0.9554.
(c) The probability that X is between 15 and 25 (inclusive) is given by = P(15 X
25) = P(X < 25.5) - P(X
14.5) {using continuity correction}
P(X < 25.5) = P( <
) = P(Z < 0.79) = 0.7852
P(X 14.5) = P(
) = P(Z
-1.70) = 1 - P(Z < 1.70)
= 1 - 0.9554 = 0.0446
The above probability is calculated by looking at the value of x = 0.79 and x = 1.70 in the z table which has an area of 0.7852 and 0.9554.
Therefore, P(15 X
25) = 0.7852 - 0.0446 = 0.7406.
Answer:
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