CIE Mathematics Paper-1: Specimen Questions 173 - 173 of 175

Question number: 173

Essay Question▾

Describe in Detail

Three points A, B and C are shown in the diagram. B is at South-East of A and South-West of C.

AC = 10 m and BC = 6 m.

A triangle showing points in different directions

A triangle showing points in different directions

Find the given calculations by using the appropriate formulas of triangle

(a) The length of AB

(b) The bearing of B and C from A.

Explanation

Here,

(a) ABC is a right angled triangle,

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

diagonal2(AC2)=Side12(AB2)+Side22(BC2) , where AC = 10 m and BC = 6 m is given,

102=AB2+62100=AB2+36AB2=10036AB2=64AB=8m .

(b) Bearing of B and C from A:

  • 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 A to another B 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 B and C from A will be NAC+θ and NAC respectively which is measured as shown below:

Triangular points showing bearing of each other

Triangular points showing bearing of each other

Showing the bearing of point B and C from point A

Now, we need to find NAC:

So, we can see that NAC+N1CA=1800 which makes straight angle.

Where, N1CA=800 is given,

So, NAC+800=1800NAC=1000

Now, tanθ=BCAB=68=0.75θ=tan10.75θ=36.870~370

  • So, bearing of B from A is NAC+θ=1000+370=1370.
  • Bearing of C from A is NAC=1000 .
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