Question

Hey guys Good Night

Plz solve this qy5​

Answer 1

Answer:

p³

Step-by-step explanation:

distance between two ponit(a, b) and (c, d) = $\sqrt{(a-c)^2+(b-d)^2}$

Now, distance between (r³ +p³cosΘ , q³ + p³sinΘ) and (r³, q³)

= $\sqrt{(r^3+p^3cos\theta - r^3)^2+(q^3+p^3sin\theta-q^3)^2}$

= $\sqrt{(p^3cos\theta)^2+(p^3sin\theta)^2}$

= $\sqrt{p^6cos^2\theta+p^6sin^2\theta}$

= $\sqrt{p^6(cos^2\theta+sin^2\theta)}$

= $\sqrt{p^6(1)}$

= p³

Answer 2

Step-by-step explanation:

distance between two ponit(a, b) and (c, d) = \sqrt{(a-c)^2+(b-d)^2}

(a−c)

2

+(b−d)

2

Now, distance between (r³ +p³cosΘ , q³ + p³sinΘ) and (r³, q³)

= \sqrt{(r^3+p^3cos\theta - r^3)^2+(q^3+p^3sin\theta-q^3)^2}

(r

3

+p

3

cosθ−r

3

)

2

+(q

3

+p

3

sinθ−q

3

)

2

= \sqrt{(p^3cos\theta)^2+(p^3sin\theta)^2}

(p

3

cosθ)

2

+(p

3

sinθ)

2

= \sqrt{p^6cos^2\theta+p^6sin^2\theta}

p

6

cos

2

θ+p

6

sin

2

θ

= \sqrt{p^6(cos^2\theta+sin^2\theta)}

p

6

(cos

2

θ+sin

2

θ)

= \sqrt{p^6(1)}

p

6

(1)

= p³