Question

If the centre of the given circle is o, find the value of x

Answer 1

(i) ∠ACB=∠ADB      (Angles in the same segment of a circle)

But ∠ADB=xo

⇒∠ABC=xo

Now in ΔABC

∠CAB+∠ABC+∠ACB=180o

⇒40o+900+xo=180o       (AC is the diameter)

⇒130o+xo=180o

⇒xo=1800−130o=50o

(ii) ∠ACD=∠ABD      (angles in the same segment)

And, ∠ACD=xo

Now in triangle OAC,

OA=OC    (radii of the same circle)

⇒∠ACO=∠AOC       (opposite angles of equal sides)

Therefore, xo=62o

(iii) ∠AOB+∠AOC+∠BOC=360o     (sum of angles at a point)

⇒∠AOB+80o+130o=360o

⇒∠AOB+210o=360o

⇒∠AOB=360

Step-by-step explanation:

As O is the centre of the circle, AB is the diameter, which can be easily seen from the figure and therefore ∠ACB is a semicircular angle. As we know Semicircular Angles are always Right angles. ∴ The value of x∘ in the following figure is 50∘.