The bearing of a point A from P is 330 degrees and the bearing of another point B from P is 290 degrees Both A and B due north of a point C which is 7600 meters due West of P. Find the distance between A and B.
This paragraph is nonsense:
"We can also use the fact that both A and B are due north of C. This means that the distance between A and C, as well as the distance between B and C, is equal to the distance between P and C, which is 7600 meters."
We cannot say that at all.
What we can say is that
BC/7600 = tan20°
AC/7600 = tan60°
That gives us
AB = AC-BC = 7600(tan60°-tan20°) = 10397 meters
You are correct, I apologize for the error in my explanation. Your solution is correct. Thank you for pointing out my mistake.
Two walkers set off at the same time from a crossroad and walk along flat straight roads inclined to each other at 68 degrees. If they both walk at a speed of 6km/hr, find their distance apart 10 minutes later.
We can start by drawing a diagram of the situation. Let's call the two walkers A and B, and let's say that they start at point O:
A
|
| 68°
|
|_____________
O B
We know that the angle between the two roads (OA and OB) is 68 degrees. We also know that both walkers walk at a speed of 6 km/hr.
After 10 minutes, A would have covered a distance of (6/60)*10 = 1 km. Similarly, B would have covered a distance of 1 km. Let's call the position of A after 10 minutes A', and the position of B after 10 minutes B':
A'
|
| 68°
|
|_____________
O B'
We want to find the distance between A' and B'. To do this, we can use the Law of Cosines. Let's call the distance between A and O "a," the distance between B and O "b," and the distance between A' and B' "c."
From the diagram, we see that angle A'OB' is 112 degrees (180 - 68 - 68). Using the Law of Cosines, we have:
c² = a² + b² - 2ab*cos(112)
c² = a² + b² + 2ab*cos(68)
We know that a = b = 1 km, so:
c² = 2 - 2*cos(68)
c = sqrt(2 - 2*cos(68))
c ≈ 1.94 km
Therefore, the distance between A' and B' after 10 minutes is approximately 1.94 km.
Two walkers set off at the same time from a crossroad and walk along flat straight roads inclined to each other at 68 degrees. If they both walk at a speed of 6km/hr, find their distance apart 10 minutes later.
We can start by drawing a diagram of the situation. Let's call the two walkers A and B, and let's say that they start at point O:
A
|
| 68°
|
|_____________
O B
We know that the angle between the two roads (OA and OB) is 68 degrees. We also know that both walkers walk at a speed of 6 km/hr.
After 10 minutes, A would have covered a distance of (6/60)*10 = 1 km. Similarly, B would have covered a distance of 1 km. Let's call the position of A after 10 minutes A', and the position of B after 10 minutes B':
A'
|
| 68°
|
|_____________
O B'
We can now use the Law of Cosines to find the distance between A' and B'. Let's call this distance "d." We know that:
- Angle A'OB' is 112 degrees (180 - 68 - 68)
- Side OA' (and OB') has length 1 km
- Both sides OA' and OB' make an angle of 68 degrees with side AB (hypothenuse of triangle AOB)
Using the Law of Cosines, we get:
d^2 = 1^2 + 1^2 - 2(1)(1)cos(112)
d^2 = 2 - 2cos(112)
We can use the fact that cos(112) = -cos(68) to simplify this expression:
d^2 = 2 + 2cos(68)
d ≈ 2.45 km
Therefore, the two walkers will be approximately 2.45 km apart after 10 minutes.
A map has a scale of 1:50000. Two towers are 10.4 km apart. How many centimeters on the map is this
We know that 1 cm on the map represents 50,000 cm (or 500 m) in real life. We want to find out how many centimeters on the map represent a distance of 10.4 km.
We can set up a proportion:
1 cm / 50,000 cm = x cm / 10,400,000 cm
where x is the number of centimeters on the map that represents a distance of 10.4 km.
We can cross-multiply:
50,000 * x = 10,400,000
x = 208 cm
Therefore, 10.4 km on the ground is represented by 208 cm on the map.
How many significant figure is 2.776
There are four significant figures in 2.776.
Two bottles of lolly water and three packers of biscuits cost 8.60. A bottle of lolly water coct 1.30 more than a packet of biscuit.
a. Find the unit cost of each item.
b. Find the total cost of five lolly water and seven packets of biscuits?
a. Let's call the cost of a pack of biscuits "x." Then, the cost of a bottle of lolly water is "x + 1.30" (because it costs 1.30 more than a pack of biscuits).
We know that two bottles of lolly water and three packs of biscuits cost 8.60. Therefore:
2(x + 1.30) + 3x = 8.60
Simplifying:
2x + 2.60 + 3x = 8.60
5x + 2.60 = 8.60
5x = 6
x = 1.20
Now we can find the cost of a bottle of lolly water:
x + 1.30 = 1.20 + 1.30 = 2.50
Therefore, a packet of biscuits costs 1.20 and a bottle of lolly water costs 2.50.
b. We want to find the total cost of five bottles of lolly water and seven packs of biscuits. Using the unit costs we just found, we have:
Total cost = (5 bottles)(2.50/bottle) + (7 packs)(1.20/pack)
Total cost = 12.50 + 8.40
Total cost = 20.90
Therefore, the total cost of five bottles of lolly water and seven packs of biscuits is 20.90.
In a herd of x cattle, the ratio of the number of bulls to cows is 1:6.
Find the number of bulls in terms of x.
If the ratio of bulls to cows is 1:6, we can write:
number of bulls : number of cows = 1 : 6
Let B be the number of bulls. Then, the number of cows must be 6B (since there are six cows for every one bull).
But the total number of cattle is x. Therefore, we can write:
B + 6B = x
Simplifying:
7B = x
So the number of bulls in terms of x is:
B = x/7
A ship has sufficient food to supply 600 passengers for 3 weeks. How long would the food last for 900 people?
We can start by using the formula:
quantity of food ∝ number of people × time
This means that if we double the number of people, the quantity of food needed doubles, and if we double the time, the quantity of food needed doubles.
Let F be the quantity of food needed for 600 passengers for 3 weeks. Then we can write:
F ∝ 600 × 3
F ∝ 1800
This means that for 600 passengers for 3 weeks, we need 1800 units of food.
To find out how long the food would last for 900 people, we can use the same formula:
F ∝ 900 × t
where t is the time in weeks.
We want to find t when F is still 1800. Using the proportion above, we get:
1800/900 = F/t
t = F/(1800/900)
t = F/2
So the food would last half as long for 900 people as it would for 600 people. Therefore, the food would last:
t = (3 weeks)/2 = 1.5 weeks
for 900 people.
Tau, Igo and Iro share a bag of mangoes in the ratio 7:6:14. If Igo received mangoes less than Iro, how many mangoes does Iro receive?
Let's say the total number of parts in the ratio is x, so we can write:
- Tau received 7 parts out of x
- Igo received 6 parts out of x
- Iro received 14 parts out of x
We can use these parts to form an equation:
7x + 6x + 14x = total number of mangoes
Simplifying:
27x = total number of mangoes
Let's say that Iro received k mangoes. Then, we know that Igo received less than k. Since the ratio of Igo to Iro is 6:14, we can write:
6/14 = Igo's share/k
Multiplying both sides by k, we get:
6k/14 = Igo's share
Simplifying:
3k/7 = Igo's share
Since Tau, Igo, and Iro share the mangoes in the ratio 7:6:14, their total share in terms of k is:
7k + 6k + 14k = 27x
Simplifying:
27k = 27x
k = x
This means that Iro received 14x parts out of x, which simplifies to:
14x/x = 14
Therefore, Iro received 14 mangoes.
The sum of three consecutive odd integers is 39. What is the smallest of these integers?
Let's call the smallest of the three consecutive odd integers "x".
Since they are consecutive odd integers, the next two odd integers must be "x + 2" and "x + 4".
The sum of these three integers is 39, so we can write:
x + (x + 2) + (x + 4) = 39
Simplifying:
3x + 6 = 39
3x = 33
x = 11
Therefore, the smallest of the three consecutive odd integers is 11.
The square of the sum of 1 and x^1/2 is equal to 25. Find the value of x.
The square of the sum of 1 and x^(1/2) can be written as:
(1 + x^(1/2))^2
Using the identity (a + b)^2 = a^2 + 2ab + b^2, we can expand the above expression:
(1 + x^(1/2))^2 = 1^2 + 2(1)(x^(1/2)) + (x^(1/2))^2
Simplifying:
(1 + x^(1/2))^2 = 1 + 2x^(1/2) + x
We are given that this expression is equal to 25, so we can write:
1 + 2x^(1/2) + x = 25
Simplifying:
x^(1/2) + x = 12
Subtracting x from both sides:
x^(1/2) = 12 - x
Now we can square both sides:
(x^(1/2))^2 = (12 - x)^2
Simplifying:
x = 144 - 24x + x^2
Rearranging:
x^2 - 25x + 144 = 0
This quadratic equation can be factored as:
(x - 9)(x - 16) = 0
Therefore, the value of x can be either 9 or 16. However, we need to check which one satisfies the original equation we were given:
(1 + x^(1/2))^2 = 25
If x = 9, then:
(1 + 3^(1/2))^2 ≈ 24.98 ≠ 25
If x = 16, then:
(1 + 4^(1/2))^2 = 25
Therefore, the value of x is 16.
Convert 116.5° to degrees, minutes and seconds
Convert the angle 116.5° to degrees, minutes and seconds
Convert Sin480° to surds form
First, we can draw a diagram to better visualize the problem:
A
|
C ------ P ------ B
We know that point C is 7600 meters due west of point P. This means that the distance between C and P is 7600 meters.
To find the distance between A and B, we can use the Law of Cosines. Let's call the distance between A and P "x," and the distance between B and P "y."
From the problem, we know that the bearing of A from P is 330 degrees. This means that the angle between the line AP and the due west line (PC) is 30 degrees. Similarly, the angle between the line BP and the due west line is 70 degrees.
Using the Law of Cosines, we have:
AB² = x² + y² - 2xy*cos(70-30)
AB² = x² + y² - 2xy*cos(40)
We can also use the fact that both A and B are due north of C. This means that the distance between A and C, as well as the distance between B and C, is equal to the distance between P and C, which is 7600 meters.
Therefore:
x + 7600 = y + 7600
x = y
Substituting this into our equation for AB², we get:
AB² = 2x² - 2x²*cos(40)
AB² = 2x²(1-cos(40))
AB = √(2x²(1-cos(40)))
Now we just need to find the value of x. To do this, we can use the fact that the bearing of A from P is 330 degrees:
tan(30) = x/7600
x = 7600*tan(30)
x = 4389.5 meters
Finally, substituting this value of x into our equation for AB, we get:
AB = √(2(4389.5)²(1-cos(40)))
AB ≈ 9907 meters
Therefore, the distance between A and B is approximately 9907 meters.