Question

please answer my question ​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{ \bf{arc(ADB) = \frac{4\pi R}{3}} }$

☆ɢɪᴠᴇɴ☆

• A circle with secant AB and tangent EBC.
• Angle $∠ABC = 120º$

✯ᴛᴏ ғɪɴᴅ✯

• Length of arc(ADB)

★ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ★

• Draw line segments MA and MB.

⍟ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ⍟

Let 'R' be the raduis of the circle.

We have,

$\to MB ⊥ EC$

$\to ∠MBC = 90º ...(1)$

Also,

$\to MB = MA$

$\to ΔMAB$ is isosceles

$\to ∠MBA = ∠MAB... (2)$

It is given that

$\to ∠ABC = 120º$

$\to ∠MBC + ∠MBA = 120º$

$\to 90^o + ∠MBA = 120º$[Using (1)]

$\to \bf{∠MBA = 30^o}$

$\to \bf{∠MAB = 30^o}$[Using (2)]

In ΔMAB,

$\to ∠MBA + ∠MAB +∠AMB = 180^o$

$\to 30^o + 30^o + ∠AMB = 180^o$

$\to ∠AMB = 180^o - 60^o$

$\to ∠AMB = 120^o$

$\to ∠AMB = 120\times \frac{\pi}{180}\:rad$

$\to ∠AMB = 120\times \frac{\pi}{180}\:rad$

$\to \bf{∠AMB = \frac{2\pi}{3}\:rad}$

Length of an arc is given by

$\to l = R\theta, \theta$ in radian.

$\to arc(AFB) = R \frac{2\pi}{3}$

$\to Circumference - arc(ADB) = \frac{2\pi R}{3}$

$\to 2\pi R - arc(ADB) = \frac{2\pi R}{3}$

$\to arc(ADB) = 2\pi R - \frac{2\pi R}{3}$

$\to arc(ADB) = \frac{(6-2)\pi R}{3}$

$\to \bf{arc(ADB) = \frac{4\pi R}{3}}$