In the process of construction of an angle with P as the 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. What will be the value of $\angle BPA$?

∠BPA = 60°.

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$∠$ BPA = $60^{\circ}$

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$∠$ BPA = $90^{\circ}$

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$∠$ BPA = $180^{\circ}$

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Solution

The correct option is B
$∠$ BPA = $60^{\circ}$

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

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Similar questions

Q. In the process of construction of an angle with P as the 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. What will be the value of

∠ B P A

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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:

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With B as the centre and any radius more than half of AB, draw another arc as shown. This is the third arc. With A as the centre and the same radius as above, cut the previously drawn arc in Y. This would the fourth arc. Join PY.

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Q. Steps for the construction of a 90° angle.

Choose the correct order.

1: Draw a ray OA.

2: With C and D as centres and radius more than half of CD, draw two arcs cutting each other at E. Join OE.

3: With O as centre and any convenient radius, draw an arc cutting OA at B.

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5: With C as centre and the same radius, draw an arc cutting again the first arc at D.

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In the process of construction of an angle with P as the 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. What will be the value of \(\angle BPA\)?

The correct option is B ∠ BPA = 60∘ Since we draw all these arcs with the same radius, it follows that PA = PB = AB. Hence △ ABP is equilateral and hence ∠ BPA = 60∘

The correct option is B
$∠$ BPA = $60^{\circ}$

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