Advertisements
Advertisements
प्रश्न
In the given figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Advertisements
उत्तर
It is given that, ABC is an equilateral triangle

We have to find `angleBDC` and `angleBEC`
Since ΔABC is an equilateral triangle
So, `angleA = angleB = angleC = 60°`
And ABEC is cyclic quadrilateral
So `angle A + angle E = 180°` (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)
Then,
`angle E = 180° - 60°`
= 120°
Similarly BECD is also cyclic quadrilateral
So,
`angle E + angle D = 180°`
`angleD = 180° - 120°`
= 60°
Hence, `angle BDC `= 60° and `angle BEC = 120°`.
APPEARS IN
संबंधित प्रश्न
In the given figure, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects PQ at A and RS at B. Prove that ∠AOB = 90º
A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
If ΔABC is isosceles with AB = AC and C (0, 2) is the in circle of the ΔABC touching BC at L, prove that L, bisects BC.
Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle.
In Fig. 4, a circle inscribed in triangle ABC touches its sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

If \[d_1 , d_2 ( d_2 > d_1 )\] be the diameters of two concentric circle s and c be the length of a chord of a circle which is tangent to the other circle , prove that\[{d_2}^2 = c^2 + {d_1}^2\].
A line segment joining any point on the circle to its center is called the _____________ of the circle
Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b.
From the figure, identify the centre of the circle.
Which of the following describes the radius of a circle?
