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 $∠ B P A$ ?

$∠$ BPA = $30^{\circ}$

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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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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 $\angle BPA$?

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

in degrees?

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.

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.

Continuing with the previous question, what angle do we have at the end of four arcs?

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.

What is the angle $∠ Y P A$ ?

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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 ∠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∘

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 $∠ B P A$ ?

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 = $60^{\circ}$