CIE Mathematics Paper-1: Specimen Questions 133 - 133 of 175

Question number: 133

Essay Question▾

Describe in Detail

A straight road between P and Q is shown in the diagram. R is the point south of P and east of Q. PR = 8.3 km and QR = 4.8 km.

A triangle PQR showing direction

A triangle PQR showing direction

Find the given calculations by using the appropriate formulas of triangle

(a) The length of the road PQ

(b) The bearing of Q from P.

Explanation

Here,

(a) PRQ is a right angled triangle,

So, by using the formula of sides of a right angled triangle:

diagonal2(PQ2)=Side12(QR2)+Side22(PR2) , where QR = 4.8 km and PR = 8.3 km is given,

PQ2=4.82+8.32=23.04+68.89=91.93PQ=9.59km .

(b) Bearing of Q from P:

  • The true bearing to a point is the angle which is measured in degrees in a clockwise direction from the north line.
  • The bearing from one point P to another Q is the direction of the line between those points.
  • The principal directions are north, east, south, and west, the pact is that the direction due north is a bearing of 0 degrees, due east is 90 degrees, due south is 180 degrees, and due west is 270 degrees.

So, here, bearing of Q from P will be 1800+θ which is measured as shown below:

triangular points in different directions

triangular points in different directions

Showing the bearing of point Q from point P

  • So, tanθ=QRPR=4.88.3=0.58θ=tan10.58θ=30.110~300
  • So, bearing of Q from P will be: 1800+θ=1800+300=2100.
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