Projection Formulae and Area of a Triangle in Trigonometry

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

Projection Formulae in Trigonometry - Projection Formulae and Area of a Triangle

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

Trigonometry area of a triangle - Projection Formulae and Area of a Triangle

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

Magnetic Field of Earth	Newton’s Law of Universal Gravitation
Magnetic Field and Magnetic Lines of Force	Variation of Resistivity with Temperature
Magnetic Properties of Materials	Newton’s First Law Of Motion
Uses of Electromagnetic Waves	Magnetism and Electromagnetism– Tamil Board