A pendulum makes 170 oscillations in 2 min 30 second what is the frequ

1.	A pendulum makes 170 oscillations in 2 min 30 second what is the frequency of the pendulum
Answer» EXPLANATION: Given: A circle with Centre O, P is the midpoint of Arc APB. PT is a tangent to the circle at P. To Prove: AB \|\| PT Construction: JOIN OA ,OB, & OP Proof: OP ⟂PT [Radius is ⟂ to tangent through point of contact] ∠OPT= 90° Since P is the midpoint of Arc APB Arc AAP =arc BP ∠AOP = ∠BOP ∠AOM= ∠BOM In ∆ AOM & ∆BOM OA= OB= r OM = OM (Common) ∠AOM= ∠BOM (PROVED above) ∠AOM≅∠BOM (by SAS CONGRUENCY axiom) ∠AMO = ∠BMO (c.p.c.t) ∠AMO + ∠BMO= 180° ∠AMO = ∠BMO= 90° ∠BMO = ∠OPT= 90° But, they are corresponding angles. Hence, AD\|\|PT Read more on Brainly.in - brainly.in/question/2013866#readmore

A pendulum makes 170 oscillations in 2 min 30 second what is the frequency of the pendulum

Answer»

EXPLANATION:

Given:

A circle with Centre O, P is the midpoint of Arc APB. PT is a tangent to the circle at P.

To Prove:

AB || PT

Construction: JOIN OA ,OB, & OP

Proof: OP ⟂PT

[Radius is ⟂ to tangent through point of contact]

∠OPT= 90°

Since P is the midpoint of Arc APB

Arc AAP =arc BP

∠AOP = ∠BOP

∠AOM= ∠BOM

In ∆ AOM & ∆BOM

OA= OB= r

OM = OM (Common)

∠AOM= ∠BOM (PROVED above)

∠AOM≅∠BOM (by SAS CONGRUENCY axiom)

∠AMO = ∠BMO (c.p.c.t)

∠AMO + ∠BMO= 180°

∠AMO = ∠BMO= 90°

∠BMO = ∠OPT= 90°

But, they are corresponding angles. Hence, AD||PT

Read more on Brainly.in - brainly.in/question/2013866#readmore

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