The geometrical interpretation of the proof of projection formulae is the length of any side of a triangle is equal to the algebraic sum of the projections of other sides upon 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: **

In any triangle ABC we have a

\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\) = 2R ……………………. (1)

Now convert the above relation into sides in terms of angles
in terms of the sides of any triangle.

a/sin A = 2R

⇒ a = 2R sin A ……………………. (2)

b/sin B = 2R

⇒ b = 2R sin B ……………………. (3)

c/sin c = 2R

⇒ c = 2R sin C ……………………. (4)

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

Now, b cos C + c cos B

= 2R sin B cos C + 2R sin C cos B

= 2R sin (B + C)

= 2R sin (π - A), [Since, A + B + C = π]

= 2R sin A

= a [From (2)]

Therefore, a = b cos C + c cos B. *Proved.*

**(ii) b = c cos A + a
cos C **

Now, c cos A + a cos C

= 2R sin C cos A + 2R sin A cos C

= 2R sin (A + C)

= 2R sin (π - B), [Since, A + B + C = π]

= 2R sin B

= b [From (3)]

Therefore, b = c cos A + a cos C.

Therefore, a = b cos C + c cos B. *Proved.*

**(iii) c = a cos B + b
cos A**

Now, a cos B + b cos A

= 2R sin A cos B + 2R sin B cos A

= 2R sin (A + B)

= 2R sin (π - C), [Since, A + B + C = π]

= 2R sin C

= c [From (4)]

Therefore, c = a cos B + b cos A.

Therefore, a = b cos C + c cos B. *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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