Question

step by step answer please​

Answer 1

$\huge\bf\blue{Tô Fíñd➽}$

(a).angle ADC

(b).angle AOC

(c).angle BAT

(d).angle OAB

________________________________

$\large\bf\blue{(a).<ADC➽}$

<ADC=180°-<ABC

➠180°-131°

➠49°......$\bf\red{Answer.}$

Reason:-

Sum of two opposite angles of quadrilateral inside the Circle is 180°

$\large\bf\orange{(b).<AOC➽}$

(SEE THE ATTACHED PIC)

<AOC=2<K (Angle formed in centre is equal to the twice angle formed in the arc if vertex are same.)

<K=180°-131°=>49°

<AOC=2×49

=>98°......$\bf\red{Answer.}$

$\large\bf\green{(a).<BAT➽}$

<BAT=<BDA

if <BDA is 20°

then,

<BAT=20°......$\bf\red{Answer.}$

REASON:-

Alternate Segment Theorem.

$\large\bf\purple{(a).<OAB➽}$

<OAT=>90°

(angle formed between tangent and radius is always 90°)

<OAB=<OAT-<BAT

<BAT=20°(Allready Prooved)

Therefore,

<OAB=90°-20°

➠70°......$\bf\red{Answer.}$

__________________________________