Question

prove projection rule
b= c cosA + a cosC
c= a cosB. + b cosA​

Answer 1

Answer:

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]

Projection Formulae

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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)

a = b cos C + c cos B

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⇒ 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

b = c cos A + a cos C

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⇒ 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.

Step-by-step explanation:

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