Trigonometry: Bearings Question?

Feeling like a bad tutor for not being able to fully solve this lol. In Q6 I’ve managed to find the triangle’s angle at the yacht corner, 92. I can’t seem to be able to find another side for the triangle in order to use cosine/sine rule. 

For Q7 the two sides given (AB from Q6 and 1.2km is all I have :/

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  • 2 months ago
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    Q6

    Draw a vertical line representing the coast, sea is to the right, land is to the left. A is down, B is up on that line. Point C representing the yacht is in the middle right. Connect the points. Angle in A is 90-32 = 58°. Angle in B is 300-270=30°

    Angle in C is 180-58-30=92°.

    We also know AB=2.2 km.

    Use Law of Sines:

    AB/sin C=BC/sin A = AC/sin B

    BC= AB sinA / sin C

    BC= 2.2 sin 58° / sin 92°= 1.867 km

    BC=1.9 km (rounded to 1 d.p.)

    AC = AB sin B / sin C

    AC = 2.2 sin 30° / sin 92°= 1.101 km

    AC = 1.1 km (rounded to 1 d.p.)

    Q7.

    BC = 1.2km

    angle B = 315-270 = 45°

    Law of Cosines

    AC^2 = AB^2 + BC^2 - 2 AB BC cos B

    AC^2 = 2.2^2 + 1.2^2 - 2*2.2*1.2*cos 45°

    AC^2 = 2.546

    AC= 1.6 (to 1 d.p.)

  • 2 months ago

    You have one side, 2.2km and one angle of 92°. This is not enough info you two angles or two sides in order to solve this problem

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