Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
Advertisements
Solution
It is given that BC || AD , `angleADC = 110°` and `angleBAC = 50°`
We have to find `angleDAC`
In cyclic quadrilateral ABCD
`angleA + angleC = 180°` ..… (1)
`angleB + angleD = 180°` ..… (2)
Since, `angleADC = 110°`
So,
`angleB = 180° - angleD`
`=180° - 110°`
= 70°
Therefore in Δ ABC , 50° + 70° + ` angle BCA `= 180°
So , `angleBCA` = 60° ..… (3)
Now, `angleBCA = angle CAD ` (BC || AD and AC is transversal)
`⇒ angle BCA = angle CAD` = 60°
APPEARS IN
RELATED QUESTIONS
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
