Question

HELLO DOSTON

ANSWER THIS PLSS
And I hate faltu answer​

Answer 1

${}{ \huge{ \tt{ \purple{Question - }}}}$

If AB is a chord of a circle with center O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.

${}{ \huge{ \tt{ \red{Solution - }}}}$

Given:-

A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at point A.

To prove:-

• ∠BAT = ∠ACB

Proof:-

∠ABC = 90° [Angle in a semicircle is a right angle]

In △ABC by angle sum property of triangle

∠ABC + ∠ BAC + ∠ACB = 180 °

∠ACB + 90° = 180° - ∠BAC

∠ACB = 90 - ∠BAC [1]

Now,

OA ⏊ AT [Tangent at a point on the circle is perpendicular to the radius through point of contact]

∠OAT = ∠CAT = 90°

∠BAC + ∠BAT = 90°

∠BAT = 90° - ∠BAC [2]

=========================================

From [1] and [2]

∠BAT = ∠ACB [Proved]

_______________________________

Answer 2

Answer:

pls aajaoo apki Yaad aati hai gheri wali bhout roya Mai aaaj pheli Baar room Mai locked rha