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.*

**The Law of Sines or The Sine Rule****Theorem on Properties of Triangle****Projection Formulae****Proof of Projection Formulae****The Law of Cosines or The Cosine Rule****Area of a Triangle****Law of Tangents****Properties of Triangle Formulae****Problems on Properties of Triangle**

**11 and 12 Grade Math**

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