prove that tan alfa =q sin theta/p+qcos theta
Answers
Answer:
Verified Answer
Step-by-step explanation:
Let θ be the angle between P and Q and R be the resultant vector. Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q.
So, we have
R = P + Q
Now, expand A to C and draw BC perpendicular to OC.
From triangle OCB,
OB² = OC² + BC²
or, OB² = (OA + AC)² + BC² ----i
In triangle ABC,
Sinθ = BC/AB
or, BC = ABSinθ
or, BC = ODSinθ {∵ AB = OD = Q}
or, BC = QSinθ
Cosθ = AC/AB
or, AC = ABCosθ
C = ABCosθ
or, AC = ODCosθ {∵ AB = OD = Q}
or, AC = QCosθ
Substituting value of AC and BC in (i), we get
R² = (P + QCosθ)² + (QSinθ)²
R² = P² + Q²Cosθ² + 2PQCosθ + Q²Sinθ²
R² = P² + Q²Cosθ² + Q²Sinθ² + 2PQCosθ
R² = P² + Q²(Cosθ² + Sinθ²) + 2PQCosθ
R² = P² + Q²+ 2PQCosθ
∴ R = √(P² + Q²+ 2PQCosθ)
which is the magnitude of resultant.
Answer of your Question starts from here.
Direction of resultant:
Let ∝ be the angle made by resultant R with P. Then,
From triangle OBC,
tan∝ = BC/OC = BC/(OA + AC)
tan∝ = QSinθ/(P + QCosθ)
which is the direction of resultant.