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If the distance between the points P(acos48,0) and Q(0,acos12)is ‘d’ then d 2 - a 2 =

If the distance between the points P(acos48,0) and Q(0,acos12)is ‘d’ then d2 - a2=

Grade:12

1 Answers

Vikas TU
14149 Points
3 years ago
Given two points in 2-dimensions:
      A = (a Cos 48° , 0)   and    B = (0, a Cos12°)
AB = d   given
=> d² = (a cos 48°  - 0)² + (0 - a Cos 12°)²
         = a² * (Cos² 48° + Cos² 12°)
         = a² * [(1 + cos 96°)/2 + (1 + cos 24°)/2 ]
         = a² * [ 2 + 2 Cos 60° * Cos 36° ] /2 
     2 d² = a² * [2 + Cos 36°] 
     2 (d² - a²) = a² Cos 36°
=>  d² - a² = a²/2 * Cos 36°
 
To find the value of Cos 36° :
A=36°Sin (2A) = Sin (180° - 3A) = Sin 3A2 Sin A Cos A = Sin A * (3 - 4 Sin² A)4 Sin² A + 2 Cos A - 3 = 04 Cos² A - 2 Cos A - 1 = 0Cos A = [1 + √5 ]/4Cos 36° = [1 + √5]/4
Given two points in 2-dimensions:
      A = (a Cos 48° , 0)   and    B = (0, a Cos12°)
AB = d   given
=> d² = (a cos 48°  - 0)² + (0 - a Cos 12°)²
         = a² * (Cos² 48° + Cos² 12°)
         = a² * [(1 + cos 96°)/2 + (1 + cos 24°)/2 ]
         = a² * [ 2 + 2 Cos 60° * Cos 36° ] /2 
     2 d² = a² * [2 + Cos 36°] 
     2 (d² - a²) = a² Cos 36°
=>  d² - a² = a²/2 * Cos 36°
 
To find the value of Cos 36° :
A=36°Sin (2A) = Sin (180° - 3A) = Sin 3A2 Sin A Cos A = Sin A * (3 - 4 Sin² A)4 Sin² A + 2 Cos A - 3 = 04 Cos² A - 2 Cos A - 1 = 0Cos A = [1 + √5 ]/4Cos 36° = [1 + √5]/4

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