5. At whats angle
do the two forces (P+Q) and (P-Q) act so
that resultant is
✓(3(P^2) + (Q^2))​

2 thoughts on “5. At whats angle<br />do the two forces (P+Q) and (P-Q) act so<br />that resultant is <br />✓(3(P^2) + (Q^2))​”

1. Given that,

   [tex]\sf:\implies\:F_1=(P+Q)[/tex]

   [tex]\sf:\implies\:F_2=(P-Q)[/tex]

   Resultant force;

   [tex]\sf:\implies\:F=\sqrt{3P^2+Q^2}[/tex]

   We have to find angle between two force vectors.

   ❖ As per parallelogram law of vector addition, magnitude of resultant vector R of two vectors A and B inclined at an angle of θ is given by;

   • R² = A² + B² + 2AB cosθ

   By substituting the given values;

   [tex]\sf\dashrightarrow\:(\sqrt{3P^2+Q^2})^2=(P+Q)^2+(P-Q)^2+2(P+Q)(P-Q)\cdot cos\theta[/tex]

   We know that,

   • (A + B)² = A² + 2AB + B²
   • (A – B)² = A² – 2AB + B²
   • (A² – B²) = (A + B) (A – B)

   [tex]\sf\dashrightarrow\:3P^2+Q^2=(P^2+2PQ+Q^2)+(P^2-2PQ+Q^2)+2(P^2-Q^2)\cdot cos\theta[/tex]

   [tex]\sf\dashrightarrow\:3P^2+Q^2=2P^2+2Q^2+2(P^2-Q^2)\cdot cos\theta[/tex]

   [tex]\sf\dashrightarrow\:(3P^2-2P^2)+(Q^2-2Q^2)=2(P^2-Q^2)\cdot cos\theta[/tex]

   [tex]\sf\dashrightarrow\:(P^2-Q^2)=2(P^2-Q^2)\cdot cos\theta[/tex]

   [tex]\sf\dashrightarrow\:2cos\theta=\dfrac{(P^2-Q^2)}{(P^2-Q^2)}[/tex]

   [tex]\sf\dashrightarrow\:cos\theta=\dfrac{1}{2}[/tex]

   [tex]\dashrightarrow\:\underline{\boxed{\bf{\orange{\theta=60^{\circ}}}}}[/tex]

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2. if the diagonals of a parallelogram are 12 cm and 15 cm find the length of each segment of the diagonal into which they are divided state the property of parallelogram used.

   please give me answer

   give answer in my Question ♡

   Reply