Question

With respect to the previous problem of construction of a ${30}^{\circ }$ angle, what angle would you get at the end of two arcs drawn in the following fashion:

With P as centre and any radius, draw an arc as shown, which cuts in A. This is the first arc. With A as centre and the same radius, cut this arc in B. This would be the second arc.

• A. ∠BPA = 60°.
• B. $\angle BPA={60}^{\circ }$
• C. $\angle BPA={90}^{\circ }$
• D. $\angle BPA={180}^{\circ }$

Answer 1

The correct option is B

$\angle BPA={60}^{\circ }$

Since we draw these arcs all with the same radius, it follows that PA = PB = AB. Hence $△ABP$ is equilateral and hence $\angle BPA={60}^{\circ }$