Proof of the projection rules

-

In triangle ABC, let us denote the magnitudes of the sides AB, BC, CA by c, a, b respectively and the angle BAC, angle CBA, angle ACB by A, B, C respectively.

Proof : 

From the cosine rules, we have

cos(B) = ((c2 + a2 - b2)/2ca), cos(C) = ((a2 + b2 - c2)/2ab)

Now, let us consider the right hand side expression in the equation to be proved.

Therefore, b cos(C) + c cos(B) = b ( (a2 + b2 - c2)/2ab ) + c ( (c2 + a2 - b2)/2ca )

⇒ b cos(C) + c cos(B) = ( (a2 + b2 - c2) + (c2 + a2 - b2) )/2a = (2a2)/2a = a.

Therefore, a = b cos(C) + c cos(B)

Similarly, we can prove that b = c cos(A) + a cos(C) and c = a cos(B) + b cos(A)

Comments

Post a Comment

Popular posts from this blog

Formula to find sum of the first n Natural numbers

-
Image
Hello, Everyone. In this, we are going to see a proof for the formula to find sum of the first n Natural numbers. Let S = 1 + 2 + 3 + 4 + ......... + (n-1) + n , where n is any natural number.  --------------- (1) The r th term from the starting of the series, in the above series is r. ------- (2) From the property of  addition of numbers, we can write S as shown below. S = n + (n-1) + (n-2) + ......... + 2 + 1 ---------- (3)  The r th term from the starting of the series, in the above series is n-(r-1) = n - r + 1 -------- (4) Let us add (1) and (3) such that the corresponding terms are added as shown in the below figure. From (2) and (4), The r th term from the starting of the series, in the resulting series obtained from addition of (1) and (3) is equal to r + (n - r + 1) = n + 1. From this, it is obvious that the r th term of the resulting series is independent of r. Therefore, as n is a constant, every term in the resulting series is equal to n + 1. Therefore, by add...
Read more

Sum of the squares of the first n natural numbers - Formula derivation

-
Image
Hello, Everyone. In this, we are going to derive the formula to find the sum of the squares of the first n natural numbers. Let S = 1 2  + 2 2  + 3 2  + .......... + (n-1) 2  + n 2 (m+1) 3 - m 3 = (m 3 + 3m 2 + 3m + 1) - m 3 = 3m 2 + 3m + 1 ------------ (i) put m = 1,2,3, ....... , (n-1), n in (i) Then, 2 3  - 1 3   = 3(1) 2  + 3(1) + 1 ------------ (1) 3 3  - 2 3   = 3(2) 2  + 3(2) + 1 ------------ (2) 4 3  - 3 3   = 3(3) 2  + 3(3) + 1 ------------ (3) ................... n 3  - (n-1) 3   = 3(n-1) 2  + 3(n-1) + 1 ------------ (n-1) (n+1) 3  - n 3   = 3(n) 2  + 3(n) + 1 ------------ (n) Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation. (n+1) 3  - 1 3   = 3(1 2  + 2 2  + 3 2  + .......... + (n-1) 2  + n 2 ) + 3(1 + 2 + 3 + ......... + (n-1) + n) + (1 + 1 + 1 +........ n times) 1 + 2 + 3 +...
Read more

Sum of cubes of the first n natural numbers - Formula derivation

-
Hello, Everyone. In this, we are going to derive the formula to find the sum of cubes of the first n natural numbers. Let S = 1 3  + 2 3  + 3 3  + .......... + (n-1) 3  + n 3 (m+1) 4  - m 4  = (m 4  + 4m 3  + 6m 2 + 4m + 1) - m 4  = 4m 3  + 6m 2 + 4m + 1 ------------ (i) put m = 1,2,3, ....... , (n-1), n in (i) Then, 2 4  - 1 4   = 4(1) 3  + 6(1) 2  + 4(1) + 1 ------------ (1) 3 4  - 2 4   = 4(2) 3  + 6(2) 2  + 4(2) + 1 ------------ (2) 4 4  - 3 4   = 4(3) 3  + 6(3) 2  + 4(3) + 1 ------------ (3) ................... n 4  - (n-1) 4   = 4(n-1) 3  + 6(n-1) 2  + 4(n-1) + 1 ------------ (n-1) (n+1) 4  - n 4   = 4(n) 3  + 6(n) 2  + 4(n) + 1 ------------ (n) Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation. (n+1) 4...
Read more