from a point o in the school compound, adeolu is 100m away on a bearing of N35 degrees east and Ibrahim is 80m away on a bearing of South 55 degrees west. how far apart are both boys and what is the bearing of Ibrahim from adeolu in three figure bearings?
Draw a diagram, and use the law of cosines to find the distance.
Then, using the point given, find the locations of the two boys. If they are at locations (a,b) and (c,d) then the bearing of Ibrahim is SθW where
tanθ = (a-c)/(b-d)
To solve this problem, we can use the concept of vectors and trigonometry.
First, let's visualize the information given. Let point O be the starting point in the school compound, with Adeolu 100m away on a bearing of N35°E, and Ibrahim 80m away on a bearing of S55°W.
To calculate the distance between Adeolu and Ibrahim, we can create two vectors: AO (from point O to Adeolu) and BO (from point O to Ibrahim). Then, we can find the length (magnitude) of the vector AB, where AB = AO + BO.
1. Calculate the components of vector AO:
- The horizontal component (west to east) is AO_hor = AO * cos(35°).
- The vertical component (south to north) is AO_ver = AO * sin(35°).
Substituting AO = 100m:
- AO_hor = 100 * cos(35°)
- AO_ver = 100 * sin(35°)
2. Calculate the components of vector BO:
- The horizontal component (east to west) is BO_hor = BO * cos(180° - 55°).
- The vertical component (north to south) is BO_ver = BO * sin(180° - 55°).
Substituting BO = 80m:
- BO_hor = 80 * cos(125°)
- BO_ver = 80 * sin(125°)
3. Calculate the components of vector AB:
- The horizontal component is AB_hor = AO_hor + BO_hor.
- The vertical component is AB_ver = AO_ver + BO_ver.
Substituting the calculated values from steps 1 and 2:
- AB_hor = AO_hor + BO_hor
- AB_ver = AO_ver + BO_ver
4. Calculate the magnitude (length) of vector AB:
- AB = √(AB_hor^2 + AB_ver^2)
Finally, to find the bearing of Ibrahim from Adeolu in three-figure bearings, we can use inverse trigonometric functions:
a. Calculate the angle theta (θ):
- θ = atan(AB_hor / AB_ver) in radian
- θ_deg = θ * (180 / π) in degrees
b. Convert the angle θ_deg to a three-figure bearing:
- If AB_hor > 0 and AB_ver > 0, the bearing will be θ_deg.
- If AB_hor > 0 and AB_ver < 0, then the bearing will be 360° + θ_deg.
- If AB_hor < 0, the bearing will be 180° + θ_deg.
In conclusion, by following these steps, we can determine the distance between both boys and the bearing of Ibrahim from Adeolu in three-figure bearings.
To find the distance between Adeolu and Ibrahim, we can use the cosine rule:
Distance squared = (Adeolu's distance)^2 + (Ibrahim's distance)^2 - 2(Adeolu's distance)(Ibrahim's distance)cos(angle between their bearings)
Angle between their bearings = 180 - (35 + 55) = 90 degrees
Distance squared = 100^2 + 80^2 - 2(100)(80)cos(90)
= 10000 + 6400 - 16000(0)
= 10000 + 6400
= 16400
Distance = √(16400) = 128.06 m (approximately)
Therefore, both boys are approximately 128.06 meters apart.
To find the bearing of Ibrahim from Adeolu, we can use the sine rule:
sine(angle between Ibrahim and Adeolu's bearing) / Ibrahim's distance = sine(Ibrahim's bearing) / distance between both boys
Sine(angle between Ibrahim and Adeolu's bearing) = sine(90) = 1 (since the angle between their bearings is 90 degrees)
sine(Ibrahim's bearing) = sine(55)
Let's calculate the distance between both boys:
Distance between both boys = √(100^2 + 80^2 - 2(100)(80)cos(90))
= √16400
= 128.06 m (approximately)
sine(55) / 128.06 = 1 / Ibrahim's distance
Ibrahim's distance = (128.06 * sine(55)) / 1
Ibrahim's distance = 105.53 m (approximately)
Therefore, the distance between Adeolu and Ibrahim is approximately 105.53 meters.
The bearing of Ibrahim from Adeolu can be calculated as follows:
Bearing of Ibrahim from Adeolu = Adeolu's bearing + 180 degrees
Bearing of Ibrahim from Adeolu = 35 + 180
Bearing of Ibrahim from Adeolu = 215 degrees.
Therefore, the bearing of Ibrahim from Adeolu is 215 degrees (three-figure bearings).