AB and BC are sides of a regular pentagon and a regular hexagon respectively. O is the common centre of both regular polygons. If OA = OB = OC, find the measure of angle ABC
Please give step by step explanation
Answers
Answer:
As given that AB is the side of pentagon the angle subtended by each arm of the pentagon at the centre of the circle is =
5
360
o
=72
o
Thus angle ∠AOB=72
o
Similarly as BC is the side of a hexagon hence the angle subtended by BC at the centre is =
6
360
o
i.e., 60
o
∠BOC=60
o
Now ∠AOC=∠AOB+∠BOC=72
o
+60
o
=132
o
The triangle thus formed, △AOB is an isosceles triangle with OA=OB as they are radii of the same circle.
Thus ∠OBA=∠BAO as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180
o
so, ∠AOB+∠OBA+∠BAO=180
o
2∠OBA+72
o
=180
o
as, ∠OBA=∠BAO
2∠OBA=180
o
−72
o
2∠OBA=108
o
∠CBA=54
o
as ∠OBA=∠BAO So,
∠OBA=∠BAO=54
o
The triangle thus formed, △BOC is an isosceles triangle with OB=OC as they are radii of the same circle.
Thus ∠OBC=∠OCB as they are opposite angles of equal sides of an isosceles triangles.
The sum of all the angles of a triangles is 180
o
so, ∠BOC+∠OBC+∠OCB=180
o
2∠OBC+60
o
=180
o
as, ∠OBC=∠OCB
2∠OBC=180
o
−60
o
2∠OBC=120
o
∠OBC=60
o
as ∠OBC=∠OCB
So, ∠OBC=∠OCB=60
o
∠ABC=∠OBA+∠OBC=54
o
+60
o
=114