Question

Prove that, S= r theta

Answer 1

Suppose the drawn circle intersects OX and OY at L and M respectively.

S is Equal to R Theta

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Clearly, arc LM subtends ∠LOM at the centre O. Now, take an arc LN of length equal to the radius of the circle and join ON.

Then, by definition, ∠LON = 1 radian.

Since the ratio of two arcs in a circle is equal to the ratio of the angles subtended by the arcs at the center of the circle, hence,

∠LOM/∠LON = arc LM/arc LN

or, ∠LOM/1 radian = arc LM/radius OL

or, ∠LOM = arc LM/radius OL × 1 radian = arc LM/ radius OL radian.

Therefore, circular measures of ∠LOM is arc LM/radius OL

If θ be the circular measure of ∠LOM, arc LM = s and radius of the circle = OL = r then,

θ = s/r, [i.e. theta equals s over r]

or, s = r θ, [i.e. s r theta formula]

Therefore, now we know the meaning of “S is equal to r theta”

Answer 2

⭐⭐HOLA⭐⭐

ANSWER:-

Let, XOY be a given angle. Now, with centre O and any radius OL draw a circle.

Clearly, arc LM subtends ∠LOM at the centre O. Now, take an arc LN of length equal to the radius of the circle and join ON.

Then, by definition, ∠LON = 1 radian.

Since the ratio of two arcs in a circle is equal to the ratio of the angles subtended by the arcs at the center of the circle, hence,

∠LOM/∠LON = arc LM/arc LN

or, ∠LOM/1 radian = arc LM/radius OL

or, ∠LOM = arc LM/radius OL × 1 radian = arc LM/ radius OL radian.

Therefore, circular measures of ∠LOM is arc LM/radius OL

If θ be the circular measure of ∠LOM, arc LM = s and radius of the circle = OL = r then,

θ = s/r, [i.e. theta equals s over r]

or, s = r θ, [i.e. s r theta formula]

Therefore, now we know the meaning of “S is equal to r theta”

hope it helps uhh

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