Given that AB = 370 m and AC = 510 m, find
(i) the distance between B and C,
(ii) ACB,
(iii) the bearing of C from B,
(iv) the shortest distance from A to BC.
Answers
Answer:
(i) The distance between B and C is approximately 552.9 m
(ii) ∠ACB is approximately 40.49°
(iii) The bearing of C from B is approximately 184.49°
(iv) The shortest distance from A to BC is approximately 331.155 m
Step-by-step explanation:
The given parameters are;
The length of segment AB = 370 m
The length of segment AC = 510 m
The bearing of B from A = 68°
The bearing of C from A = 144°
(i) From the bearing of B from A = 68° and the bearing of C from A = 144°, we have;
∠BAC = 144° - 68° = 76°
∠BAC = 76°
Let A represent ∠BAC, c represent segment AB, b represent segment AC, and a represent segment BC, by cosine rule, we have;
a² = b² + c² - 2·b·c·cos(A)
By substituting the known values, we get;
a² = 510² + 370² - 2 × 510 × 370 × cos(76°)
a² ≈ 397000 - 91301.3234 ≈ 305698.677
BC = a ≈ √(305,698.677) ≈ 552.9
The distance between B and C, BC ≈ 552.9 m
(ii) By sine rule, we have;
a/(sin(A) = b/(sin(B)) - c/(sin(C))
Therefore;
552.9/(sin (76°)) = 370/(sin(C))
sin(C) = 370/(552.9/(sin (76°))) ≈ 0.649321
C = arcsine(0.649321) ≈ 40.49°
∠C = ∠ACB ≈ 40.49°
(iii) Angle ∠B = ∠ABC = 180° - 76° - 40.49° ≈ 63.51°
The bearing of A from B = 360° - (180° - 68°) = 248°
Therefore, the bearing of C from B ≈ 248° - 63.51° ≈ 184.49°
(iv) The shortest distance from A to BC = 370 m × sin(63.51°) ≈ 331.155 m
Related Questions
The mean and standard deviation of the sample data collected on continuous variable are -0.25 and 0.03, respectively. The following table shows the
relative frequencies of the data in the given intervals.
Interval
Relative Frequency
0.02
0.15
0.33
-0.34 < x < -0.31
-0.31 <<< -0.28
-0.28 << -0.25
-0.25 < 3 < -0.22
-0.22 <3 < -0.19
-0.19 <<< -0.16
0.36
0.11
0.03
Based on the table, do the data support the use of a normal model to approximate population characteristics?
A Yes, because the sum of the relative frequencies is 1.00.
B
Yes, because the distribution of relative frequencies is very close to the empirical rule for normal models.
с
No, because the values are negative and normal models are used for positive values.
D
No, because the distribution of relative frequencies is very far from the empirical rule for normal models.
E
No, because the sample size and the population parameters are not known.
Answers
Answer:
B
Step-by-step explanation:
I know because I know that how
The correct answer is D. No, because the distribution of relative frequencies is very far from the empirical rule for normal models.
What is Mean?The sum of all values divided by the total number of values determines the mean (also known as the arithmetic mean, which differs from the geometric mean) of a dataset. The term "average" is frequently used to describe this measure of central tendency.
To determine if the data supports the use of a normal model to approximate population characteristics, we need to examine the relative frequencies and see if they exhibit a bell-shaped or normal distribution.
Looking at the table, we see that the relative frequencies are not symmetrical and do not follow the empirical rule for normal models.
The largest relative frequency is in the interval from -0.28 to -0.25, and the frequencies decrease as we move away from this interval. Additionally, there are negative values in the data, which is not a characteristic of normal models.
Therefore, the correct answer is D. No, because the distribution of relative frequencies is very far from the empirical rule for normal models.
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Answer:
Step-by-step explanation:
SOH CAH TOA to help remember how the trig functions fit on the triangle
use SOH since we know the angle and the hypotenuse and want to find the opposite side
Sin(x)= Opp/ Hyp
Hyp* sin(38) = Opp
12* 0.6156614753 = Opp
7.3879377...
7.4 is the answer
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Answers
Answer:
i know im late but i think the answer is A.
Step-by-step explanation:
im sorry if im wrong but its the most reasonable answer.
Answer:
A
Step-by-step explanation:
PLEASE ANSWER< WILL GIVE BRAINLY TO BEST ANSWER!?
Answers
Answer: D is the correct answer.
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Answers
Answer:
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Step-by-step explanation:
hope this helps
Aisha wants to paint the four walls of her living room.
Each wall is 2.4 m high and 3.7 m long.
One wall has a door of 2 m by 0.7 m.
Tins of paint cost £12 per 1.5 L tin.
Each litre of paint can cover 8 m2 of wall.
There is an offer of: Buy 2 tins get the 3rd at half price.
How much will Aisha pay to paint her living room?
Answers
9514 1404 393
Answer:
£30
Step-by-step explanation:
The total area of the four walls is ...
A = 4 × LW
A = 4 × (2.4 m)(3.7 m) = 35.52 m²
Subtracting the area of the door, we find the painted area to be ...
35.52 m² -(2 m)(0.7 m) = 34.12 m²
Painting this area will require ...
(34.12 m²)/(8 m²/L × 1.5 L/tin) = 2.8433... tins
Aisha will need to buy 3 tins of paint, for which she pays 2.5 times the price of 1 tin:
2.5 × £12 = £30 . . . . Aisha's cost to paint her living room
£25 or £20.10
Step-by-step explanation:
calculate the area to be painted:
2.1x4.5= 9.45m^2 for each wall
x4= 37.8m^2
area of door= 1.8x0.9= 1.62m^2
subtract the area of the door from the wall area
37.8-1.62= 36.18m^2 to be painted
divide the area by the area that 1L of paint can cover
16.18/12= 3.015 L needed
if 1.5L= £10 then, 3L= £20. she needs 2 cans + an extra 0.015 L- so another can (3 cans total) add price up.
£30- but buy 2 get 3rd 1/2 price, so take £5 off. price is £25
or
calculate price for 0.015L
1.5=£10
divide both sides by 100
0.015= £ 0.10
if 1.5L= £10 then, 3L= £20. she needs 2 cans + an extra 0.015 L
£20+£0.10= £20.10
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Answers
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Step-by-step explanation:
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Answers
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Answers
Answer/Step-by-step explanation:
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Step-by-step explanation:
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