Question

side AB of quadrilateral is diameter of its inner circle & angle ADC=140° Then find angle BAC​

Answer 1

GivEn:

• Side AB of quadrilateral is diameter of its inner circle.

• $\sf \angle ADC = 140^\circ$

To find:

• $\sf \angle BAC = 140^\circ$

SoluTion:

ABCD is a cyclic quadrilateral.

So, As we know that,

In a cyclic quadrilateral the sum of each pair of opposite angles is $\sf 180^\circ$.

Therefore,

$:\implies\sf \angle ADC + \angle ABC = 180^\circ$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 140^\circ + \angle ABC = 180^\circ$

⠀⠀⠀

$:\implies\sf \angle ABC = 180^\circ - 140^\circ$

⠀⠀⠀

$:\implies\bf \angle ABC = 40^\circ$

━━━━━━━━━━━━━━━

AB is diameter of Circle.

Therefore,

$:\implies\sf \angle ACB = 90^\circ\;\;\;\;\bigg\lgroup\bf \because\;Angle\;in\; semicircle \bigg\rgroup$

⠀⠀⠀

★ Now, In ∆ ABC

⠀⠀⠀

$:\implies\sf \angle ABC + \angle ACB + \angle BAC = 180^\circ\;\;\;\;\bigg\lgroup\bf \because\;Angle\;sum\; property\;of\; Triangle \bigg\rgroup$

⠀⠀⠀

$:\implies\sf 40^\circ + 90^\circ + \angle BAC = 180^\circ$

⠀⠀⠀

$:\implies\sf 130^\circ + \angle BAC = 180^\circ$

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$:\implies\sf \angle BAC = 180^\circ - 130^\circ$

⠀⠀⠀

$:\implies{\underline{\boxed{\sf{\pink{\angle BAC = 50^\circ}}}}}\;\bigstar$

⠀⠀⠀

$\therefore\;\sf \underline{Hence,\;\angle BAC\:is\; \bf{50^\circ.}}$

Answer 2

Answer :

$\Large\bold{\underline{\mathtt{\pink{Given:}}}}$

» Side AB of quadrilateral is diameter of its inner circle .

» $\sf \angle ADC\:=\:140°$

$\Large\bold{\underline{\mathtt{\pink{To\:Find:}}}}$

» $\sf \angle BAC$

$\Large\bold{\underline{\mathtt{\pink{Sølution:}}}}$

» $\sf \angle ACB\:=\:90°$ (angle in a semicircle)

Now , In cyclic quadrilateral ABCD ,

• The sum of opposite angles is 180° .

➡ $\sf \angle ADC\:+\: \angle ABC\:=\:180°$

➡ $\sf 140°\:+\: \angle ABC\:=\:180°$

➡ $\sf \angle ABC\:=\:180°\:-\:140°$

➡ $\large\bold{\boxed{\bf{\red{ \angle ABC\:=\:40°}}}}$

Now , Angle Sum Property of Triangle is 180° .

➡ $\sf \angle BCA\:+\: \angle ABC\:+\: \angle BAC\:=\:180°$

➡ $\sf 90°\:+\: 40°\:+\: \angle BAC\:=\:180°$

➡ $\sf 130°\:+\: \angle BAC\:=\:180°$

➡ $\sf \angle BAC\:=\:180°\:-\:130°$

➡ $\large\bold{\boxed{\bf{\red{ \angle BAC\:=\:50°}}}}$

• For figure , refer to the attachment .

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