## The Cosine Formulae:

In △ABC with BC = a, CA = b and AB = c,

(i) cos A = (b^{2} + c^{2} – a^{2})/2bc (ii) cos B = (c^{2} + a^{2} – b^{2})/2ca (iii) cos C = (a^{2} + b^{2} – c^{2})/2ab

**Proof-** The proofs of all three formulae are similar. Let us take the first one. We consider three different types of △ABC- (i) acute-angled (ii) right-angled at A and (iii) obtuse-angled at A.

**Case (i):**

CD is drawn perpendicular to AB, In △CAD, CD/AC = sin A [Fig. (i)] ∴ CD = AC sin A = b sin A Now, DB = c – AD = c – (AD/AC)AC = c – b cos A Now, in right-angled △CDB, BC ^{2} = CD^{2} + BD^{2}⇒ a ^{2} = (b sin A)^{2} + (c – b cos A)^{2}⇒ a ^{2} = b^{2} sin^{2} A + c^{2} – 2bc cos A + b^{2} cos^{2} A⇒ a ^{2} = b^{2} (sin^{2} A + cos^{2} A) + c^{2} – 2 bc cos A⇒ a ^{2} = b^{2} + c^{2} – 2 bc cos A∴ cos A = (b ^{2} + c^{2} – a^{2})/2bc |

**Case (ii):**

In △ABC, right-angled at A [Fig. (ii)] a ^{2} = b^{2} + c^{2} = b^{2} + c^{2} – 2bc cos A (∵ cos A = cos 90° = 0)∴ 2 bc cos A = b ^{2} + c^{2} – a^{2}∴ cos A = (b ^{2} + c^{2} – a^{2})/2bc |

**Case (iii):**

In △ABC, ∠A > 90°. CD is drawn perpendicular to BA produced [Fig. (iii)] ∴ In right-angled triangle CDA, CD/AC = sin A, ∴ CD = AC sin A = b sin A Also, AD/AC = cos A, ∴ AD = AC cos A = b cos A Now, in right-angled triangle CDB, BC ^{2} = CD^{2} + DB^{2}∴ a ^{2} = (b sin A)^{2} + (b cos A + c)^{2}⇒ a ^{2} = b^{2} sin^{2} A + b^{2} cos^{2} A + 2bc cos A + c^{2}⇒ a ^{2} = b^{2} + c^{2} + 2bc cos A∴ 2bc cos A = b ^{2} + c^{2} – a^{2}∴ cos A = (b ^{2} + c^{2} – a^{2})/2bc |

From the above three cosine formulae, one can write-

a^{2} = b^{2} + c^{2} – 2bc cos Ab ^{2} = c^{2} + a^{2} – 2ca cos Bc ^{2} = a^{2} + b^{2} – 2ab cos C |

Adding up, a^{2} + b^{2} + c^{2} = 2 (a^{2} + b^{2} + c^{2}) – 2 (bc cos A + ca cos B + ab cos C)

⇒ a^{2} + b^{2} + c^{2} = 2 (bc cos A + ca cos B + ab cos C)

As a^{2} + b^{2} + c^{2} > 0, bc cos A + ca cos B + ab cos C > 0