Question

In the given figure, the radius of the circle AB = AC = 5 cm and BC is a chord. Find

its distance from the centre, if Angle ABC = 60.​

Answer 1

Answer:

$\frac{5\sqrt{3} }{2}$

Step-by-step explanation:

AB+AC=5cm[radius]

Angle ABC= Angle ACB[angle opposite to equal side]

It distance = height od the triangle

you can use sin∅ or cos∅

The height/Side=sin60°

h/5=√3/2

H=$\frac{5\sqrt{3} }{2}$

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Answer 2

Best Answer you would ever find for this one:

AB = AC = 5cm

->Clearly, A is the Centre of the circle

∠ABC = 60°

AB = AC therefore ∠ABC = ∠ACB

∠ACB = 60

Now,

∠ABC+∠ACB+∠CAB = 180°

∠CAB = (180-120)°

∠CAB = 60°

Draw AM ⊥BC

⇒CM = BM = 1/2 x BC

(Perpendicular from the center to a chord bisects the chord)

∠ABC = ∠ACB = ∠CAB = 60°

⇒ ΔABC IS an equilateral Δ

    ⇒Bc = 5 cm

CM = 1/2 x BC

CM = 2.5 cm

AM = √AC²-CM²

AM = √5²-(2.5)²

After some calculation

AM = $\sqrt\frac{75}{4}$ = 2.5√3

Hope its helpful