Advertisements
Advertisements
प्रश्न
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Advertisements
उत्तर
\[ \stackrel\frown{QP} \text{ is a major arc and } \angle PSQ \text{ is the angle formed by it in the alternate segment } . \]
\[ \text{ We know that the angle subtended by an arc at the centre is twice the angle subtended by it at any point of the alternate segment of the circle } . \]
`=> 2angle "PSQ" = "m"`
`=> 2angle "PSQ" = 360^circ - "m"`
`=> 2 angle"PSQ" = 360^circ - 180^circ ...(because angle "POQ" < 108^circ)`
`=> 2angle "PSQ" > 180^circ`
`=> angle "PSQ" > 90^circ`
Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.
APPEARS IN
संबंधित प्रश्न
Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
Given an arc of a circle, complete the circle.
In the below fig. O is the centre of the circle. Find ∠BAC.
If O is the centre of the circle, find the value of x in the following figure
If O is the centre of the circle, find the value of x in the following figure
If O is the centre of the circle, find the value of x in the following figures.
In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
In the given figure, two circles intersect at A and B. The centre of the smaller circle is Oand it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
In the given figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
In the following figure, ∠ACB = 40º. Find ∠OAB.
