Advertisements
Advertisements
प्रश्न
In the given figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.
Advertisements
उत्तर
It is given that, `angle BAC = 60°` and `angle BCA = 20°`
We have to find the `angle ADC `
In given Δ ABC we have
`angle ABC + angle BCA + angle BAC `= 180° \[ \left( \text{ Angle sum property } \right)\]
\[ \Rightarrow \angle ABC = 180° - \left( 60° + 20° \right) = 100° \]
In cyclic quadrilateral ABCD we have
`angle B + angleD = 180°` (Sum of pair of opposite angles of a cyclic quadilateral is 180º)
Then,
`angle D = 180° - 100° `
`angle D = 80° `
Hence `angle ADC = 80°`
APPEARS IN
संबंधित प्रश्न
In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = `\frac { 1 }{ 2 } ("perimeter of ∆ABC")`
O is the center of a circle of radius 8cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB
If from any point on the common chord of two intersecting circles, tangents be drawn to circles, prove that they are equal.
In the given figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
In a cyclic quadrilateral ABCD, if m ∠A = 3 (m ∠C). Find m ∠A.
In the given figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is
Use the figure given below to fill in the blank:
R is the _______ of the circle.
Use the figure given below to fill in the blank:
______ is a chord of the circle.
Draw circle with the radii given below.
4 cm
The circumcentre of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90º.
