Village Q is 7km from village P on a bearing of 060. VIllage R is 5 Km from village P on a bearing of 315. Using scale drawing with a scale of 1 cm : 1 km.
find:
a. the direct distance from village Q to village R
b. the bearing of village Q from village R.
From where 105 come?
I think 105 came from 315-180=135 and then 135-(90-60)
Where does the 1.2 and 4.8 come from
Well, aren't those villages having fun with their bearings and distances! Let's dive into it:
a. To find the direct distance from village Q to village R, we can use the Pythagorean theorem. The horizontal distance between Q and R is equal to the difference in their easting coordinates, which is 7 km - 5 km = 2 km. The vertical distance is equal to the difference in their northing coordinates, which is 0 km - 5 km = -5 km (negative because R is below P). So, we have a right-angled triangle with sides of 2 km and 5 km. Using the Pythagorean theorem (a^2 + b^2 = c^2), we can find the direct distance (c) between Q and R:
c = sqrt((2 km)^2 + (-5 km)^2)
c ≈ 5.39 km
b. To find the bearing of village Q from village R, we can use some trigonometry. The bearing is the angle measured clockwise from the north direction. We know that village R is 5 km from P on a bearing of 315 degrees. To determine the bearing of Q from R, we need to find the angle between the line connecting Q and R, and the line connecting R and P. Let's take a look at it:
P(060)
/ \
/ \
Q ____________ R(315)
To find the bearing of Q from R, we need to find the angle QPR. Since R is 315 degrees from P and Q is 60 degrees from P, the angle QPR is equal to:
QPR = |315 degrees - 60 degrees|
QPR = 255 degrees
So, the bearing of village Q from village R is 255 degrees (measured clockwise from the north direction).
Hope that puts a smile on your face! If you have any more questions, fire away!
To find the answers, we can use trigonometry and the concept of vectors.
a. The direct distance from village Q to village R can be found by adding the vectors representing the distances and directions from village P to village Q and village P to village R.
First, let's represent each distance and direction as vectors. The vector from village P to village Q can be represented as (7 km, 060°), and the vector from village P to village R can be represented as (5 km, 315°).
To add the vectors, we need to break them down into their horizontal and vertical components.
For the vector (7 km, 060°), the horizontal component is given by 7 km * cos(60°) = 7/2 km. The vertical component is given by 7 km * sin(60°) = (7√3)/2 km.
For the vector (5 km, 315°), the horizontal component is given by 5 km * cos(315°) = -(5√2)/2 km. The negative sign indicates a direction opposite to the positive x-axis. The vertical component is given by 5 km * sin(315°) = -(5√2)/2 km, again with a negative sign.
Now, we can add the horizontal and vertical components of the vectors separately.
Horizontal component: (7/2 km) + (-(5√2)/2 km) = (7 - 5√2)/2 km
Vertical component: (7√3)/2 km + (-(5√2)/2 km) = (7√3 - 5√2)/2 km
Using the Pythagorean theorem, the magnitude (or direct distance) of the resulting vector is given by:
Magnitude = √(horizontal component^2 + vertical component^2)
Plugging in the values, we have:
Magnitude = √((7 - 5√2)/2 km)^2 + ((7√3 - 5√2)/2 km)^2)
Now, we can simplify and calculate the direct distance from village Q to village R using the given scale of 1 cm : 1 km.
b. To find the bearing of village Q from village R, we can use trigonometry and the concept of angles.
The bearing is the angle measured clockwise from the north direction to the line connecting village R to village Q. To calculate the bearing, we can use the tangent function.
Using the horizontal and vertical components of the vector from village R to village Q, which are (-7/2 km) and (-7√3)/2 km respectively, we can find the bearing.
Bearing = atan2((horizontal component), (vertical component))
Plugging in the values, we have:
Bearing = atan2((-7/2 km), (-7√3)/2 km)
Now, we can calculate the bearing using a scientific calculator or use trigonometric tables.
Note: Make sure to convert the radians result to degrees if required.
angle QPR is 105°
so, using the law of cosines, the distance
QR^2 = 7^2+5^2-2*7*5*cos105°
QR = 9.6 km
Expressing the vectors PQ and PR as complex numbers relative to P,
RQ = 7cis60°-5cis105° = 4.8+1.2i
so the angle θ of RQ has
tanθ = 1.2/4.8
θ = 14°
Thus, the bearing of Q from R is 90-14 = 76°