Flashcards · 7 cards

Sine & cosine rules

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  1. Rule

    How do you label a triangle for the sine and cosine rules?

    Answer

    Side aa faces angle AA, side bb faces angle BB, side cc faces angle CC.

    ABCabc
    Sides and opposite anglesSide a faces angle A, and so on
  2. Know it

    The sine rule?

    Answer

    asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}.

  3. Know it

    The cosine rule?

    Answer

    c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C. For an angle: cos⁡C=a2+b2−c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}.

  4. Know it

    When do you use the sine rule, and when the cosine rule?

    Answer

    Sine rule: you know a side and the angle opposite it. Cosine rule: two sides and the angle between them, or all three sides.

  5. Know it

    What is cos⁡120∘\cos 120^\circ, and what does it do in the cosine rule?

    Answer

    −12-\frac12. So −2abcos⁡120∘=+ab-2ab\cos 120^\circ = +ab: the side opposite an obtuse angle comes out longer.

  6. Rule

    The bisector of angle MM in triangle MNOMNO meets NONO at PP. How does it divide NONO?

    Answer

    In the ratio of the other two sides: NPPO=MNMO\dfrac{NP}{PO} = \dfrac{MN}{MO}.

    MNOPNP : PO = MN : MO
    The angle bisectorThe longer side next to M gets the longer piece
  7. Which method?

    WAEC 2017 · Paper 2 · Q9 (a)

    An aeroplane flies 100 km from town AA on a bearing of 330∘330^\circ to town BB. It then flies 300 km due west to town CC. (i) Illustrate this information in a diagram. (ii) Calculate the: (I) distance between AA and CC, correct to two decimal places; (II) bearing of CC from AA.

    How do you find the angle at BB?

    Answer

    At BB, the bearing back to AA is 330∘−180∘=150∘330^\circ - 180^\circ = 150^\circ and on to CC is 270∘270^\circ, so ∠ABC=120∘\angle ABC = 120^\circ. Then the cosine rule gives ACAC.