Mathematics
In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.
Answer
Since, AB = AC.
Hence, in △ABC,
∠ACB = ∠ABC = 50°.
Since sum of angles in a triangle = 180°.
In △ABC,
⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 50° + 50° + ∠BAC = 180°
⇒ ∠BAC + 100° = 180°
⇒ ∠BAC = 180° - 100° = 80°.
From figure,
∠BDC = ∠BAC = 80° (∵ angles in same segment are equal.)
In cyclic quadrilateral sum of opposite angles = 180°,
Hence in BDCE,
⇒ ∠BDC + ∠BEC = 180°
⇒ 80° + ∠BEC = 180°
⇒ ∠BEC = 180° - 80° = 100°.
Hence, the value of ∠BDC = 80° and ∠BEC = 100°.
Related Questions
In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral are produced to meet at a point P and the sides AD and BC produced to meet at a point Q. If ∠ADC = 75° and ∠BPC = 50°, find ∠BAD and ∠CQD.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
In the figure (i) given below, ABDC is a cyclic quadrilateral. If AB = CD, prove that AD = BC.
