Question

In the given figure, the radius of the circle with centre O is 6 cm. find the length of AC.​

Answer 1

Answer:

11.59 cm

Or

$6 \times \sqrt{2 + \sqrt{3} }$

Step-by-step explanation:

We know that

OA = OC = OB = 6 cm

and

<AOB = 2× <ACB (from the theorem: Angle subtended by a point in circumference is half of angle subtended by centre for a same arc)

Therefore, <ACB = 30°

and <ACO = <BCO = 15° (because AD= DB)

Now in ∆AOC

It is an isosceles triangle as OA=OC

Therefore, <OAC = <OCA = 15°.

As the sum of all angles of a triangle is 180°.

Therefore <AOC = 180° - 15° - 15° = 150°

And after that cosine is applied.

kindly check attachment to see diagram and further steps.

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ length \: of \: AC. \\ \\ so \: here \: in \\ △ \: AOB \\ ∠ AOB = 60 \\ \\ so \: thus \: then \\ ∵ \: AO=OB \\ \\ ∠ABO = ∠BAO = 60$

$so \: hence \: then \: \\ △AOB \: \: is \: an \: equilateral \: triangle \\ \\ so \: then \\ AB = AO = OB = 6.cm \\ \\ so \: then \: here \\ OD \: ⊥ \: AB \: \: \: \: \: \: \: \: (given)\\ \\ considering \: △ODA \\ ∠ODA = 90 \\ ∠OAD = 60 \\ ∴ \: ∠AOD = 30 \\ \\ hence \: here \\ △ODA \: is \: a \\ \: 30 - 60 - 90 \: \: triangle$

$we \: know \: that \\ side \: opp. \: to \: 60 = \frac{ \sqrt{3} }{2} \times hypotenuse \\ \\ OD = \frac{ \sqrt{3} }{2} \times AO \\ \\ = \frac{ \sqrt{3} }{2} \times 6 \\ \\ AO= 3 \sqrt{3} .cm$

$so \: then \\ by \: observation \: \\ we \: can \: say \: that \\ ∠AOC \: is \: an \: obtuse \: angled \: triangle \\ \\ so \: hence \: then \\ its \: longest \: altitude \: ie \: its \: \\ longest \: perpendicular \\ lies \: in \: its \: exterior \: ie \: AD \\ \\ so \: moreover \: here \\ OC \: is \: another \: circumradius \: of \\ given \: circle \\ OC = 6.cm$

$now \: by \: applying \\ obtuse - angled \: \: triangle \: \\ containing \: \: property \\ we \: get \\ \\ AC {}^{2} = OC {}^{2} + AO {}^{2} + 2.OC \times OD \\ \\ = (6) {}^{2} + (6) {}^{2} +( 2 \times 6 \times 3 \sqrt{3} ) \\ \\ = 36 + 36 + 36 \sqrt{3} \\ \\ = 72 + 36 \sqrt{3} \\ \\ = 36(2 + \sqrt{3} \: ) \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: have \: \\ \\ AC = 6 \sqrt{2 + \sqrt{3} } \: cm$