Advertisements
Advertisements
Question
Prove that the angle in a segment greater than a semi-circle is less than a right angle.
Advertisements
Solution

\[\text{ To prove } : \angle ABC \text{ is an acute angle } \]
\[\text{ Proof } : \]
\[\text{ AD being the diameter of the given circle } \]
\[ \Rightarrow \angle ACD = 90° \left[ \text{ Angle in a semicircle is a right angle } \right]\]
\[\text{ Now, in } \bigtriangleup ACD, \angle ACD = 90° \text{ which means that } \angle ADC \text{ is an acute angle } . . . . . . \left( 1 \right)\]
\[\text{ Again, } \angle ABC = \angle ADC \left[ \text{ Angle in a same segment are always equal } \right]\]
\[ \Rightarrow \angle ABC \text{ is also an acute angle } . \left[ \text{ Using } \left( 1 \right) \right]\]
\[\text{ Hence proved } \]
APPEARS IN
RELATED QUESTIONS
Given an arc of a circle, show how to complete the circle.
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 figure

O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.
In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.

In the given figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.

In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =

In the following figure, ∠ACB = 40º. Find ∠OAB.

