Question

prove that tan alfa =q sin theta/p+qcos theta
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Answer 1

Answer:

Verified Answer

Answer 2

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.