Question

23° B In the given figure, the side BC of angle ABC has been produced on both the sides; on the left to D and on the right to E. If angle ABD = 96° and angle ACH = 130°, find angle ABC, angke ACB-and angle BAC.​

Answer 1

Step-by-step explanation:

$\huge\fbox\pink{Answer}$

$In\: this\: figure\: ABD+ABC=180° {linear\: pair}$

$=>96°+ABC=180°$

$=>ABC=180°-96°$

$=>ABC=84°$

$And \:ACE+ACB=180° {linear \:pair}$

$=>130°+ACB=180°$

$=>ACB=180°-130°$

$=>ACB=50°$

$BAC+ABC+ACB=180°.$ ${sum \:of \:angle \:of \:a \:triangle}$

$BAC+84°+50°=180°$

$BAC=180°-(84°+50°)$

$BAC=180°-134°$

$BAC= 46°$

Answer 2

★ In this figure ABD+ABC= 180° linearpair

= > 96°+ABC=180°

= > ABC=180°-96°

= > ABC=84°

★ And ACE + ACB = 180 ° Linear Pair

= > 130°+ACB=180°

= > ACB=180°-130°

= > ACB=50°=>ACB=50°

★BAC+ABC+ACB= 180°

Sum Of Angle of A Triangle

BAC+84°+50°=180°

BAC=180°-(84°+50°)

BAC=180°-134°

BAC= 46°