Question

EVALUATE---
1 ) R = √P² + Q² + 2PQ cos 0°
2 ) R = √P² + Q² + 2PQ cos 90°
PLEASE EXPLAIN BRIEFLY-------​

Answer 1

Explanation:

1)R=√P² + Q² + 2PQ cos 0°

but cos 0°=1

so R=√P² + Q² + 2PQ

R=√(P+Q)^2

R=P+Q

2)R = √P² + Q² + 2PQ cos 90°

but cos 90°=0

R = √P² + Q² + 2PQ×0

R = √P² + Q²

Answer 2

To find the magnitude of the resultant -

Draw a perpendicular CE in AB, after producing it.

From right angled triangle AEC,

AC² = AE² + CE²

= (AB + BE)² + CE²

= AB² + BE² + 2AB.BE + CE²

• (a+b)² = a² + b² + 2ab.

= AB² + BC² + 2AB. BE

• BE² + CE² = BC²

From right angled triangle BEC,

BE for Base/BC for hypotenuse = Cos.

• BE = BCCos.

Therefore, AC² = AB² + BC² + 2AB.BCCos.

Now, Let AC = R, AB = P, BC = Q.

R² = P² + Q² + 2PQCos.

R = √P² + √Q² + √2PQ. Cos.

Now, special cases.

1) If Cos = 0°

R = √P² + √Q² + √2PQ

Since, Cos0° = 1,

R = √(P + Q)² → (a+b)² formula.

or, R = P + Q.

2) If Cos = 90°

R = √P² + √Q² + √2PQ

Since, Cos90° = 0,

R = √(P - Q)² → (a-b)² formula.

or, R = P - Q.