Table of Contents

## Projection Formulae and Area of a Triangle:

### Projection Formulae:

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

- a = b cos C + c cos B
- b = c cos A + a cos C
- c = a cos B + b cos A

The proofs of all three formulae are similar. Let us establish the first relation. As the triangle may be acute-angled, right-angled or obtuse-angled, we consider ∠C to be acute, right-angled or obtuse.

Case (i) When C < 90°. AD is drawn perpendicular on BC [Fig: (i)]a = BC = BD + DC = AB (BD/AB) + AC (DC/AC) ⇒ a = c cos B + b cos C |

Case (ii) When C = 90° [Fig: (ii)]a = BC = AB (BC/AB) = c cos B + b cos 90° (∵ cos 90° = 0) ⇒ a = c cos B + b cos C |

Case (iii) When C > 90°. AD is drawn perpendicular to BC produced [Fig (iii)]Here, a = BC = BD – CD = AB (BD/AB) – AC (CD/AC) ⇒ a = c cos B – b cos (180° – C) ⇒ a = c cos B + b cos C |

The other two formulae can similarly be proved. The above three formulae are known as projection Formulae.

### Area of a Triangle:

The area of △ABC is given by

△ = (1/2) ab sin C

△ = (1/2) bc sin A

△ = (1/2) ca sin B

**Proof:**

Case (i) When △ABC is acute-angled [Fig (i)]△ = (1/2) base x altitude = (1/2) BC x AD = (1/2) ab sin C |

Case (ii) When △ABC is right-angled [Fig (ii)]△ = (1/2) BC x AC [∵ AC ⊥ BC] ⇒ △ = (1/2) ab = (1/2) ab sin C [∵ sin C = sin 90° = 1] |

Case (iii) When △ABC is obtuse-angled [Fig (iii)]△ = (1/2) base x altitude = (1/2) BC x AD ⇒ △ = (1/2) a AC sin (180° – C) [∵ ∠ACD = 180° – C] ⇒ △ = (1/2) ab sin C |

The other two formulae for △ can also be proved similarly.

**Example-** In any triangle ABC, prove that a^{3} cos (B – C) + b^{3} cos (C – A) + c^{3} cos (A – B) = 3 abc.

**Solution:**

a cos (B – C) = k sin A cos (B – C) ⇒ a cos (B – C) = k sin (B + C) cos (B – C) ⇒ a cos (B – C) = (1/2) k 2 sin (B + C) cos (B – C) ⇒ a cos (B – C) = (1/2) k (sin 2B + sin 2C) ⇒ a cos (B – C) = k (sin B cos B + sin C cos C) ⇒ a cos (B – C) = k sin B cos B + k sin C cos C ⇒ a cos (B – C) = b cos B + c cos C Similarly, b cos (C – A) = c cos C + a cos A and c cos (A – B) = a cos A + b cos B Thus, L.H.S. = a ^{2} (b cos B + c cos C) + b^{2} (c cos C + a cos A) + c^{2} (a cos A + b cos B)⇒ L.H.S. =ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C) ⇒ L.H.S. = ab . c + bc . a + ca . b = 3abc |