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Precalculus Honors · Study guide

2.4 · Right and oblique triangles: choosing a law

Match the known triangle configuration to an efficient law, and check whether SSA data gives zero, one, or two triangles.

Precalculus Honors · Reading: Demana 10th ed. §§4.2, 4.8, 5.5–5.6; Stewart 8th ed. §§6.2, 6.5–6.6

Learning objectives

  • Use right-triangle ratios with correct side labels.
  • Choose the Law of Sines or Law of Cosines.
  • Check ambiguous SSA data rather than assuming uniqueness.

Match sides with their opposite angles

Use lowercase a,b,c\displaystyle a,b,c opposite angles A,B,C\displaystyle A,B,C. In a right triangle, sine is opposite over hypotenuse and cosine is adjacent over hypotenuse. A diagram makes the reference angle explicit.

For nonright triangles, a known opposite side-angle pair suggests the Law of Sines. SSS or SAS data suggests the Law of Cosines.

Use the appropriate oblique-triangle law

The Law of Sines is a/sin⁡A=b/sin⁡B=c/sin⁡C\displaystyle a/\sin A=b/\sin B=c/\sin C. The Law of Cosines is c2=a2+b2−2abcos⁡C\displaystyle c^2=a^2+b^2-2ab\cos C. The included angle in SAS belongs between the two known sides.

Triangle area can be found by K=absin⁡C/2\displaystyle K=ab\sin C/2 using two sides and their included angle. Keep enough precision in intermediate values if a decimal answer is required.

c2=a2+b2−2abcos⁡Cc^2=a^2+b^2-2ab\cos C

SSA may have two solutions

An inverse sine gives one angle, but its supplement may also fit. After finding B\displaystyle B, check whether 180∘−B\displaystyle 180^\circ-B is another candidate with a positive remaining angle.

If the computed sine exceeds 1\displaystyle 1, no triangle exists. If two candidates both satisfy the angle-sum condition, there are two triangles. Do not use SSA as a congruence rule.

Worked examples

Use SAS data

Two sides have lengths 5\displaystyle 5 and 7\displaystyle 7 with included angle 60∘\displaystyle 60^\circ. Find the third side.

  1. Use the Law of Cosines.
  2. The square is 25+49−70(1/2)=39\displaystyle 25+49-70(1/2)=39.
  3. The side length is 39\displaystyle \sqrt{39}.

Use an opposite pair

If A=30∘\displaystyle A=30^\circ, B=45∘\displaystyle B=45^\circ, and a=4\displaystyle a=4, find b\displaystyle b.

  1. The Law of Sines gives b/sin⁡45∘=4/sin⁡30∘\displaystyle b/\sin45^\circ=4/\sin30^\circ.
  2. The ratio on the right is 8\displaystyle 8.
  3. Thus b=42\displaystyle b=4\sqrt{2}.

Common mistakes

  • Pairing a side with a nonopposite angle in the Law of Sines.
  • Using the wrong angle as the included angle.
  • Ignoring the supplementary SSA possibility.

Try it yourself

No calculator required.

1. Two sides are 5\displaystyle 5 and 7\displaystyle 7 with included angle 60∘\displaystyle 60^\circ. Find the third side and the area.

A triangle has sides 5 and 7 enclosing a 60 degree angle.
The marked angle lies between the two known sides.
Show worked solution

The third side is 39\displaystyle \sqrt{39} by the Law of Cosines. The area is (1/2)(5)(7)sin⁡60∘=353/4\displaystyle (1/2)(5)(7)\sin60^\circ=35\sqrt{3}/4.

A triangle has sides 5 and 7 enclosing a 60 degree angle. Law of Cosines gives the third side as square root of 39.
Law of Cosines gives the third side as square root of 39.

2. If A=30∘\displaystyle A=30^\circ, a=4\displaystyle a=4, and b=6\displaystyle b=6, what are the possible values of angle B\displaystyle B?

Show worked solution

The Law of Sines gives sin⁡B=3/4\displaystyle \sin B=3/4. The candidates are arcsin⁡(3/4)\displaystyle \arcsin(3/4) and 180∘−arcsin⁡(3/4)\displaystyle 180^\circ-\arcsin(3/4). Both leave a positive third angle because arcsin⁡(3/4)>30∘\displaystyle \arcsin(3/4)>30^\circ, so two triangles exist.

3. A right triangle has hypotenuse 10\displaystyle 10 and angle 30∘\displaystyle 30^\circ. Find the opposite and adjacent legs.

Show worked solution

The opposite leg is 10sin⁡30∘=5\displaystyle 10\sin30^\circ=5. The adjacent leg is 10cos⁡30∘=53\displaystyle 10\cos30^\circ=5\sqrt{3}.

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