Find the distribution of
R=Asinθ
, where A is a fixed constant and θ is uniformly distributed on
(−π/2,π/2)
. Such a random variable R arises in the theory of ballistics. If a projectile is fired from the origin at an angle α from the earth with a speed v, then the point R at which it returns to the earth can be expressed as
R=(v2/g)sin2α
, where g is the gravitational constant, equal to 980 centimeters per second squared..

The distribution of R=Asinθ is P(θ<arcsin (t/A)) = (arcsin (t/A))π+ ½

Step-by-step explanation:

Given

To find the distribution of R=Asinθ

θ is uniformly distributed on (−π/2,π/2)

R=(v²/g)sin2α

g = 980 cm/s²

Since R=Asinθ

Divide Through by A

R/A = Asinθ/A

R/A = sinθ

So, P(R<t) = P(Asinθ<t) this is equivalent to

P(sinθ<t/A)

For t > A; P(sinθ<t/A) = 1

For |t| ≤ A; P(sinθ<t/A) = P(θ<arcsin (t/A))

For t < A; P(sinθ<t/A) = 0

So, we have

P(sinθ<t/A) = P(θ<arcsin (t/A))

P(θ<arcsin (t/A)) = (arcsin (t/A) + π/2)/π

P(θ<arcsin (t/A)) = (arcsin (t/A))/π + (π/2)/π

P(θ<arcsin (t/A)) = (arcsin (t/A))π+ ½

Hence, the distribution of R=Asinθ is P(θ<arcsin (t/A)) = (arcsin (t/A))π+ ½

The distribution of the random variable R is derived as;

[(1/π)(sin^(-1) (t/A))] + (1/2)

Step-by-step explanation:

From the question,

Random variable, R=Asinθ

So, the distribution of this variable can be represented as;

P(R < t) = P(A sin θ < t)

So let's apply conditions to it to obtain;

P(R < t) = P(A sin θ < t) {1, P(θ < sin^(-1) (t/A)), 0

At conditions of, t > A ; |t| ≤ A ; t < A; respectively.

Therefore, since uniformly distributed on (−π/2,π/2), we finally arrive at;

P(θ < sin^(-1) (t/A)) = [(sin^(-1) (t/A)) + (π/2)]/π

= [(1/π)(sin^(-1) (t/A))] + (1/2)

answer: 45

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true