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 a3 cos (B – C) + b3 cos (C – A) + c3 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. = a2 (b cos B + c cos C) + b2 (c cos C + a cos A) + c2 (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 |
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