Well, when a cyclist is negotiating a circular path, there are two force which act on him 1 is the weight (W) which act vertically download and the second is the normal reaction (N) which acts perpendicular to the inclined path. and the horizontal component of the normal reaction provide the necessary centripetal force which is required to keep the cyclist moving in a circular. path. So your equations for vertical equilibrium and horizontal equilibrium volume. would be Vertical equilibrium: Ncosθ = mg Horizontal equilibrium (centripetal force): Nsinθ = mv²/r Solving for the Radius: Dividing the second equation by the first, we get: tanθ = v²/rg Dividing the second equation by the first, we get: tanθ = v²/rg Substituting the given values: tan(30°) = (10)² / (10 * r) 1/√3 = 100 / (10r) r = 100√3 / 10 r = 10√3 meters Therefore, the radius of the circular path is 10√3 meters.
Question : $\angle A, \angle B$ and $\angle C$ are three angles of a triangle. If $\angle A- \angle B=15^{\circ}, \angle B - \angle C=30^{\circ}$, then $\angle A, \angle B$ and $\angle C$ are:
Option 1: $80^{\circ},60^{\circ},$ and $40^{\circ}$
Option 2: $70^{\circ},50^{\circ},$ and $60^{\circ}$
Option 3: $80^{\circ},65^{\circ},$ and $35^{\circ}$
Option 4: $80^{\circ},55^{\circ},$ and $45^{\circ}$
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