prove that : a) angle in the major segment of a circle is acute b) angle in the minor segment of circle is obtuse
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PSQ is major segment. and PRQ is minor segment in the circle.
we have to prove: angle prq is acute and psq is obtuse.
* Draw op, oq and pq.
We know that-- angle suspended by arc of circle at center is twice the angle suspended at any point on remaining part of the circle.
so, ∠ poq =2∠prq
but , 2∠ poq <180° (∵ ∠ of Δpoq)
∠poq<90°. ----> acute angle.
-----------------------------------------------------------------------------------------------------
now reflex angle poq = 2∠psq
but ∠ poq >180°.
so 2∠psq >180°
∴∠psq>90° ---->obtuse angle.
we have to prove: angle prq is acute and psq is obtuse.
* Draw op, oq and pq.
We know that-- angle suspended by arc of circle at center is twice the angle suspended at any point on remaining part of the circle.
so, ∠ poq =2∠prq
but , 2∠ poq <180° (∵ ∠ of Δpoq)
∠poq<90°. ----> acute angle.
-----------------------------------------------------------------------------------------------------
now reflex angle poq = 2∠psq
but ∠ poq >180°.
so 2∠psq >180°
∴∠psq>90° ---->obtuse angle.
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Answer:
Step-by-step explanation:
PSQ is major segment. and PRQ is minor segment in the circle.
we have to prove: angle prq is acute and psq is obtuse.
* Draw op, oq and pq.
We know that-- angle suspended by arc of circle at center is twice the angle suspended at any point on remaining part of the circle.
so, ∠ poq =2∠prq
but , 2∠ poq <180° (∵ ∠ of Δpoq)
∠poq<90°. ----> acute angle.
-----------------------------------------------------------------------------------------------------
now reflex angle poq = 2∠psq
but ∠ poq >180°.
so 2∠psq >180°
∴∠psq>90° ---->obtuse angle.
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