The Cosine Rule Flashcards

All 8 cards in this deck

State the cosine rule for finding a missing side.

a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A

AA is the angle between sides bb and cc, opposite side aa.

State the cosine rule rearranged to find a missing angle.

cos⁡A=b2+c2−a22bc\cos A=\dfrac{b^2+c^2-a^2}{2bc}

What are the steps to find a side using the cosine rule?

e.g. sides 88 cm and 1111 cm with an included angle of 72∘72^\circ

  1. Substitute: a2=82+112−2×8×11×cos⁡72∘a^2=8^2+11^2-2\times 8\times 11\times\cos 72^\circ
  2. Correct order of operations: 64+121−54.4=130.664+121-54.4=130.6
  3. Square root: a=11.4a=11.4 cm (3 s.f.)

What are the steps to find an angle using the cosine rule?

e.g. find the angle opposite the 77 cm side in a triangle with sides 55 cm, 88 cm, 77 cm

  1. Substitute into cos⁡A=b2+c2−a22bc\cos A=\dfrac{b^2+c^2-a^2}{2bc}: cos⁡A=52+82−722×5×8\cos A=\dfrac{5^2+8^2-7^2}{2\times5\times8}
  2. Simplify: cos⁡A=4080=0.5\cos A=\dfrac{40}{80}=0.5
  3. Inverse cosine: A=60∘A=60^\circ

True or false? In a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A you may work out b2+c2−2bcb^2+c^2-2bc first and then multiply by cos⁡A\cos A.

False.

You must multiply 2bc2bc by cos⁡A\cos A first, then subtract that from b2+c2b^2+c^2.

What are the steps to use the cosine rule in a bearings problem?

e.g. a ship sails 88 km on a bearing of 060∘060^\circ, then 55 km on a bearing of 150∘150^\circ; find its distance from the start

  1. Sketch the journey with a north line at each turning point.
  2. Use angle facts to find the interior angle of the triangle: 90∘90^\circ
  3. Apply the cosine rule: d2=82+52−2×8×5cos⁡90∘d^2=8^2+5^2-2\times8\times5\cos 90^\circ, so d=9.43d=9.43 km (3 s.f.)

True or false? If the cosine rule gives a negative value for cos⁡A\cos A, then angle AA is obtuse.

True. cos⁡\cos is negative for angles between 90∘90^\circ and 180∘180^\circ, so AA is obtuse.

What are the steps to find the distance from the start of a two-leg journey given as bearings?

e.g. a boat sails 55 km on a bearing of 050∘050^\circ, then 77 km on a bearing of 110∘110^\circ

  1. Sketch the journey and find the angle inside the triangle at the turning point: 180∘−(110∘−50∘)=120∘180^\circ-(110^\circ-50^\circ)=120^\circ
  2. Two sides and the included angle, so use the cosine rule: d2=52+72−2×5×7×cos⁡120∘=109d^2=5^2+7^2-2\times5\times7\times\cos120^\circ=109
  3. Square root: d≈10.4d\approx10.4 km