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Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
Answer:
(0, 8/3) = y-intercept
Step-by-step explanation:
I chose to find the y-intercept since it is easy to find with the given information.
The equation for a line is y = mx + b
To find the y-intercept, we must plug in the info we have and solve for b:
4 = -1/3 (-4) + b
4 = -4/3 + b
8/3 = b
To find a point on the line, you also could have added -1 to 4 and 3 to -4 since slope means rise/run or change in y/change in x:
(y - 1) / (x + 3) = (4 - 1) / (-4 + 3) = (3, -1) (flip since it's the y and x coordinate) = (-1, 3)
Eight below zero would express the value -8, negative eight,
but not 8.
