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Geometry and Trigonometry - Cosine Rule

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Cosine Rule is used to relate the side lengths and angles of any triangle (not just right-angled ones). It is essentially a generalization of the Pythagorean theorem. In any triangle ABCABC with side lengths a,b,ca, b, c opposite to angles A,B,CA, B, C, the rule provides a way to find a third side if two sides and the included angle (SAS) are known.

Standard labeling of a triangle with sides a, b, c opposite angles A, B, C.
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To find an unknown side, use the standard form: a2=b2+c2−2bccos⁡(A)a^2 = b^2 + c^2 - 2bc \cos(A). This is applicable when you have the lengths of two sides and the size of the angle trapped between them.

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To find an unknown angle, use the rearranged form: cos⁡(A)=b2+c2−a22bc\cos(A) = \frac{b^2 + c^2 - a^2}{2bc}. This is applicable when the lengths of all three sides (SSS) are known.

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When calculating angles, if cos⁡(A)\cos(A) is negative, the angle AA is obtuse (between 90∘90^\circ and 180∘180^\circ). The calculator's cos⁡−1\cos^{-1} function will automatically provide the correct obtuse angle.

📐Formulae

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

b2=a2+c2−2accos⁡(B)b^2 = a^2 + c^2 - 2ac \cos(B)

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

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

cos⁡(B)=a2+c2−b22ac\cos(B) = \frac{a^2 + c^2 - b^2}{2ac}

cos⁡(C)=a2+b2−c22ab\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}

💡Examples

Problem 1:

In triangle ABCABC, side b=7 cmb = 7 \text{ cm}, side c=10 cmc = 10 \text{ cm}, and ∠A=52∘\angle A = 52^\circ. Calculate the length of side aa correct to 3 significant figures.

Solution:

Using the Cosine Rule: a2=b2+c2−2bccos⁡(A)a^2 = b^2 + c^2 - 2bc \cos(A) a2=72+102−2(7)(10)cos⁡(52∘)a^2 = 7^2 + 10^2 - 2(7)(10) \cos(52^\circ) a2=49+100−140cos⁡(52∘)a^2 = 49 + 100 - 140 \cos(52^\circ) a2=149−140(0.6156...)a^2 = 149 - 140(0.6156...) a2=149−86.19...a^2 = 149 - 86.19... a2=62.80...a^2 = 62.80... a=62.80...≈7.92 cma = \sqrt{62.80...} \approx 7.92 \text{ cm}

Explanation:

We identify this as a Side-Angle-Side (SAS) problem. We substitute the known values into the side-finding version of the formula and take the square root to find aa.

Problem 2:

A triangle has sides of length 5 cm5 \text{ cm}, 8 cm8 \text{ cm}, and 11 cm11 \text{ cm}. Find the size of the largest angle in the triangle.

Solution:

The largest angle is opposite the longest side. Let a=11a = 11, b=5b = 5, and c=8c = 8. We need to find ∠A\angle A: cos⁡(A)=b2+c2−a22bc\cos(A) = \frac{b^2 + c^2 - a^2}{2bc} cos⁡(A)=52+82−1122(5)(8)\cos(A) = \frac{5^2 + 8^2 - 11^2}{2(5)(8)} cos⁡(A)=25+64−12180\cos(A) = \frac{25 + 64 - 121}{80} cos⁡(A)=−3280=−0.4\cos(A) = \frac{-32}{80} = -0.4 A=cos⁡−1(−0.4)≈113.6∘A = \cos^{-1}(-0.4) \approx 113.6^\circ

Explanation:

Since all three sides are known (SSS), we use the rearranged formula to find the angle. The longest side (11) must be aa to find the largest angle AA.

Problem 3:

A surveyor measures two sides of a triangular plot of land. Side xx is 45 m45\text{ m} and side yy is 60 m60\text{ m}. The angle between these two sides is 70∘70^\circ. Find the length of the third side zz to 2 decimal places.

Triangle with sides 45m and 60m and included angle 70 degrees.

Solution:

  1. Identify the given values: x=45x = 45, y=60y = 60, and included angle θ=70∘\theta = 70^\circ.
  2. Apply the Cosine Rule formula for the side length: z2=452+602−2(45)(60)cos⁡(70∘)z^2 = 45^2 + 60^2 - 2(45)(60) \cos(70^\circ)
  3. Calculate the squares and product: z2=2025+3600−5400(0.3420...)z^2 = 2025 + 3600 - 5400(0.3420...) z2=5625−1846.93...z^2 = 5625 - 1846.93... z2=3778.06...z^2 = 3778.06...
  4. Take the square root: z=3778.06...≈61.47 mz = \sqrt{3778.06...} \approx 61.47\text{ m}

Explanation:

Since we have two sides and the included angle (SAS), we use the side-length version of the Cosine Rule to find the side opposite the given angle.

Problem 4:

In triangle PQRPQR, the side lengths are PQ=6 cmPQ = 6\text{ cm}, QR=4 cmQR = 4\text{ cm}, and PR=9 cmPR = 9\text{ cm}. Find the size of the angle QQ to the nearest tenth of a degree.

Obtuse triangle with sides 4, 6, and 9 cm labeled.

Solution:

  1. Let p=4p = 4 (side opposite PP), q=9q = 9 (side opposite QQ), and r=6r = 6 (side opposite RR).
  2. We need to find angle QQ, so use the rearranged Cosine Rule: cos⁡(Q)=p2+r2−q22pr\cos(Q) = \frac{p^2 + r^2 - q^2}{2pr}
  3. Substitute the values: cos⁡(Q)=42+62−922(4)(6)\cos(Q) = \frac{4^2 + 6^2 - 9^2}{2(4)(6)} cos⁡(Q)=16+36−8148\cos(Q) = \frac{16 + 36 - 81}{48} cos⁡(Q)=−2948\cos(Q) = \frac{-29}{48}
  4. Calculate the inverse cosine: Q=cos⁡−1(−2948)Q = \cos^{-1}\left(-\frac{29}{48}\right) Q≈127.168...∘Q \approx 127.168...^\circ
  5. Rounding to 1 decimal place, Q=127.2∘Q = 127.2^\circ.

Explanation:

With three sides known (SSS), the Cosine Rule allows us to solve for any angle. A negative cosine value correctly indicates that the angle is obtuse.