Derive the law and cosines and law of sines by using triangle law and vector addition
Answers
= It states that if two vectors acting simultaneously at a point are represented in magnitude and direction by the two sides of a triangle taken in same order. And their resultant is represented in magnitude and direction by the third side of the triangle taken in opposite order.
→ [Diagram is in attachment]
Consider two vectors A vector and B vector represented by OP and PQ. Let the angle between A vector and B vector is Q (theta) by the two sides of a triangle. Resultant to be OD vector by third side of triangle taken in opposite order. Draw DN perpendicular to OP produced.
In ∆ OND (By Pythagoras)
(R)² = (ON)² + (ND)²
(R)² = (OP + PN)² + (ND)²
(R)² = (A + PN)² + (NQ)² ..............(S)
In ∆ PDN
PN ÷ PD = Cos Q
PN ÷ B = Cos Q
PN = B Cos Q ..........(1)
ND ÷ PQ = Sin Q
ND ÷ B = Sin Q
ND = B Sin Q .............(2)
Put value of (1) and (2) in (S)
(R)² = (A + B Cos Q)² + (B Sin Q)²
(R)² = A² + B² Cos²Q + 2AB Cos Q + B² Sin² Q
R = √A² + B² (Sin²Q + Cos²Q) + 2AB CosQ
R = √A² + B² + 2AB Cos Q
Let R vector make an angle Π with A vector.
tan Π = DN ÷ ON
= B Sin Q ÷ OP + PN
= B Sin Q ÷ A + B Cos Q
