An equilateral triangle of side 9 cm
Answers
Answer:
plz mark me brainliest
Step-by-step explanation:
△ABC is an equilateral triangle
AB=BC=CA=9cm
O is the circumcentre of △ABC
∴OD id the perpendicular of the side BC
In △OBD and △ODC
OB=OC (Radius of the circle)
BD=DC (D is the mid point of BC)
OD=OD (common)
∴△OBD=△ODC
⇒∠BOD=∠COD
∠BOC=2∠BAC=2×60
∘ =120
(The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle)
∴∠BOD=∠COD= 2
∠BOC = 2
120 =60
BD=BC= 2
BC = 2 9 cm
In △BOD
⇒sin∠BOD=sin60
= OB , BD
Answer:
Step-by-step explanation:
△ABC is an equilateral triangle
AB=BC=CA=9cm
O is the circumcentre of △ABC
∴OD id the perpendicular of the side BC
In △OBD and △ODC
OB=OC (Radius of the circle)
BD=DC (D is the mid point of BC)
OD=OD (common)
∴△OBD=△ODC
⇒∠BOD=∠COD
∠BOC=2∠BAC=2×60
∘ =120
(The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle)
∴∠BOD=∠COD= 2
∠BOC = 2
120 =60
BD=BC= 2
BC = 2 9 cm
In △BOD
⇒sin∠BOD=sin60
= OB , BD