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.
11 and 12 Grade Math
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