Answer:
Part A
The bearing of the point 'R' from 'S' is 225°
Part B
The bearing from R to Q is approximately 293.2°
Step-by-step explanation:
The location of the point 'Q' = 35 km due East of P
The location of the point 'S' = 15 km due West of P
The location of the 'R' = 15 km due south of 'P'
Part A
To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that
=
, the right triangle ΔRPS is an isosceles right triangle
∴ ∠PRS = ∠PSR = 45°
The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°
Part B
∠PRQ = arctan(35/15) ≈ 66.8°
Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°
Answer:
A. (1,0)
Step-by-step explanation:
When you're solving a system of equations algebraically, you're finding the point that exists on both functions. So, when you're finding the solution graphically, you just need to find the point where the two functions intersect. In this case, it's A. (1,0)
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Answer:
<u>option (2) it is and isosceles triangle. </u>
Step-by-step explanation:
In circle L, points E and F lie on the circle such that E, F and L are not collinear
If LE, LF and EF are drawn
So, LE = EF = the radius of the circle L
So, LEF is an isosceles triangle.
The answer is option (2) it is and isosceles triangle.
Answer:
B
Step-by-step explanation: