# Projection Formulae

Projection formulae is the length of any side of a triangle is equal to the sum of the projections of other two sides on it.

In Any Triangle ABC,

(i) a = b cos C + c cos B

(ii)  b = c cos A + a cos C

(iii) c = a cos B +  b cos A

Proof:

Let ABC be a triangle. Then the following three cases arises:

Case I: If ABC is an acute-angled triangle then we get,

a = BC = BD + CD ………………………… (i)

 Now from the triangle ABD we have,cos B = BD/AB    ⇒ BD = AB cos B ⇒ BD = c cos B, [since, AB = c] Again, cos C = CD/AC   ⇒ CD = AC cos C  ⇒ CD = b cos C, [since, AC = b]

Now, substitute the value of BD and CD in equation (i) we get,

a = c cos B + b cos C

Note: We observe in the above diagram BD and CD are projections of AB and AC respectively on BC.

Case II: If ABC is an acute-angled triangle then we get,

a = BC = CD - BD                                           ………………………… (ii)

 Now from the triangle ADC we have, cos C =  CD/AC    ⇒ CD = AC cos C ⇒ CD = b cos C, [since, AC = b] Again, cos (π - B) = BD/AB   ⇒ BD = AB cos (π - B)

⇒ BD = -c cos B, [since, AB = c and cos (π - θ) = -cos θ]

Now, substitute the value of BD and CD in equation (ii) we get,

a = b cos C - (-c cos B)

⇒ a = b cos C + c cos B

Case III: If ABC is a right-angled triangle then we get,

a = BC                                              ………………………… (iii)

 and cos B =  BC/AB    ⇒ BC = AB cos B ⇒ BC = c cos B, [since, AB = c] Now, substitute the value of BC in equation (iii) we get,        a = c cos B ⇒ a = c cos B + 0

⇒ a = c cos B + b cos C, [since C = 90° ⇒ cos C = cos 90 = 0]

Therefore, in any triangle ABC we get, a = b cos C + c cos B

Similarly, we can prove that the formulae b = c cos A + a cos C and c = a cos B + b cos A.

Solved problem using the projection formulae:

If the area of the triangle ABC be ∆, show that,
b$$^{2}$$ sin 2C + c$$^{2}$$ sin 2B = 4∆.

Solution:

b$$^{2}$$ sin 2C + c$$^{2}$$ sin 2B

= b$$^{2}$$ 2 sin C cos C + c$$^{2}$$ ∙ 2 sin B cos B

= 2b cos C ∙ b sin C + 2c cos B ∙ c sin B

= 2b cos C ∙ c sin B + 2c cos B ∙ c sin B, [Since, a/sin B = c/sin C b sin C = c sin B]

= 2c sin B (b cos C + c cos B)

= 2c sin B ∙ a [Since, we know that, a = b cos C + c cos B]

= 4 ∙ ½ ac sin B

= 4∆.                        Proved.

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