e Find the area of the figure.
ii) In the figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is
produced to E. If <ADE = 700, and <OBA = 45°, find <BAC and <OCA.
Answers
Answer:
Here AED will not be 95° , ADE is 95°
solution ÷ Using linear pair therom,
ADE + ADC = 180°
or, ADC = 180°- ADE = 180°-95°
or, ADC = 85°
Now, because ABCD is a cyclic quadrilateral
so,
ABC + ADC= 180°
or, (ABO +OBC)+ADC= 180°
or, ( 30° +OBC) +95°= 180°
or, OBC= 180°- 125°= 65°
As, BO and CO are radii of same circle
so, BO=CO
so, OBC=OCB [ angle opposite to equal
opposite sides are equal]
In Triangle OBC , by sum property
OBC+ OCB + BOC=180°
by putting all the values
BOC= 50°
or, 2BAC = 50° [we know angle
substanded by an arc
at the centre is twice
the angle substanded
by it at remaining circle]
or, BAC=25°
Again,
In Triangle AOB
AO= BO [ Radii of same circle]
so, OAB=OBA (=30°)
Finally,
OAC= OAB - BAC= 30°-25°=5°
Hence,
OAC=5° ANS
Answer:
∠ OCA = 20° and ∠ BAC = 65°.
Step-by-step explanation:
∠ ADE + ∠ ADC = 180 (linear pair)
But, ∠ ADC = 70° (given)
Therefore,
∠ ADE + 70 = 180
∠ ADE = 180 - 70 = 110
Note that ABCD is a cyclic quadrilateral. (Quadrilateral inscribed in a circle).
In a cyclic quadrilateral, opposite angles are supplementary.
Therefore,
∠ ABC + ∠ ADC = 180
∠ ABC + 110 = 180
∠ ABC = 70
The reflex angle ∠ AOC is an angle subtended by the arc ADC at the center and ∠ ABC is the angle made by the same arc at a point B on the arc other than ACD.
Therefore,
∠ AOC = 2 ∠ ABC
= 2 × 70
= 140
Note that OA = OB = OC (radii)
Therefore,
Δ OAB, Δ OBC, and Δ OCA are all isosceles.
Now, in Δ OAB,
OA = OB and therefore,
∠ OAB = ∠ OBA (angles opposite to equal sides are equal)
Since ∠ OBA = 45° (given),
∠ OAB = 45°
In triangle Δ AOC,
∠ OAC = ∠ OCA and ∠ AOC = 140
By angle-sum property,
∠ AOC + ∠ OAC + ∠ OCA = 180°
140 + 2 ∠ OCA = 180°
2 ∠ OCA = 180 - 140 = 40
∠ OCA = 20 and ∠ OAC = 20
∠ BAC = ∠ OAB + ∠ OAC
= 45 + 20
= 65
Hence, ∠ OCA = 20° and ∠ BAC = 65°.