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A person travels from point E to point F, a distance of 4 km on a bearing of 135°. Then, the person travels from point F to point G, a distance of 13 km on a bearing of 45°. Draw a diagram of the jour

Mathematics

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4. (a) Illustrate the information in a diagram.

A diagram showing the journey:

  • Start at point E. Draw a North line vertically upwards from E.
  • From E, draw a line segment EF, 4 km long, at an angle of 135∘135^\circ clockwise from the North line. Point F is the end of this segment.
  • From point F, draw another North line vertically upwards, parallel to the first North line.
  • From F, draw a line segment FG, 13 km long, at an angle of 045∘045^\circ clockwise from the North line at F. Point G is the end of this segment.
  • Connect points E and G to form triangle EFG.

The angle ∠EFG\angle EFG can be determined from the bearings. The bearing of F from E is 135∘135^\circ. The angle between the North line at F and the line FE (pointing towards E) is 180∘−135∘=45∘180^\circ - 135^\circ = 45^\circ (alternate interior angles with the 135∘135^\circ bearing from E, considering the line from F to E). The bearing of G from F is 045∘045^\circ. Therefore, the interior angle ∠EFG=45∘+45∘=90∘\angle EFG = 45^\circ + 45^\circ = 90^\circ.

4. (b) i. Find ∣EG∣|EG| correct to two significant figures.

  1. Identify the sides and angle of triangle EFG.

    We have ∣EF∣=4|EF| = 4 km, ∣FG∣=13|FG| = 13 km, and ∠EFG=90∘\angle EFG = 90^\circ.

  2. Use the cosine rule to find $|EG|$.

    Since ∠EFG=90∘\angle EFG = 90^\circ, we can use the Pythagorean theorem.

    ∣EG∣2=∣EF∣2+∣FG∣2−2∣EF∣∣FG∣cos⁡(∠EFG)|EG|^2 = |EF|^2 + |FG|^2 - 2|EF||FG|\cos(\angle EFG) ∣EG∣2=42+132−2(4)(13)cos⁡(90∘)|EG|^2 = 4^2 + 13^2 - 2(4)(13)\cos(90^\circ) ∣EG∣2=16+169−2(4)(13)(0)|EG|^2 = 16 + 169 - 2(4)(13)(0) ∣EG∣2=185|EG|^2 = 185

  3. Calculate $|EG|$ and round to two significant figures.

    ∣EG∣=185≈13.60147|EG| = \sqrt{185} \approx 13.60147

    Correct to two significant figures, ∣EG∣|EG| is 14 km\boxed{\mathbf{14 \text{ km}}}.

    4. (b) ii. The bearing of G from F.

  4. Read the information directly from the problem statement.

    The problem states: "He continues 13km on a bearing of 045° to G." This means the bearing of G from F is 045∘045^\circ.
    The bearing of G from F is 045∘\boxed{\mathbf{045^\circ}}.

    5. (a) The midpoint of the line joining Y(2a,3)Y(2a, 3) and Z(−4,b)Z(-4, b) is M(1,2a−1)M(1, 2a-1). Find the values of aa and bb.

  5. Use the midpoint formula for the x-coordinate.

    The x-coordinate of the midpoint is Mx=xY+xZ2M_x = \frac{x_Y + x_Z}{2}.

    1=2a+(−4)21 = \frac{2a + (-4)}{2} 2=2a−42 = 2a - 4 6=2a6 = 2a a=3a = 3

  6. Use the midpoint formula for the y-coordinate.

    The y-coordinate of the midpoint is My=yY+yZ2M_y = \frac{y_Y + y_Z}{2}.

    2a−1=3+b22a - 1 = \frac{3 + b}{2}

  7. Substitute the value of aa found in Step 1 into the equation from Step 2 and solve for bb.

    2(3)−1=3+b22(3) - 1 = \frac{3 + b}{2} 6−1=3+b26 - 1 = \frac{3 + b}{2} 5=3+b25 = \frac{3 + b}{2} 10=3+b10 = 3 + b b=7b = 7

    The values are a=3,b=7\boxed{\mathbf{a = 3, b = 7}}.

    5. (b) Find ∣YZ∣|YZ|.

  8. Determine the coordinates of points Y and Z using the values of aa and bb.

    Y(2a,3)=Y(2(3),3)=Y(6,3)Y(2a, 3) = Y(2(3), 3) = Y(6, 3)
    $Z

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