Question

find the x in the following figures​

Answer 1

$\large\underline{\underline\mathrm\red{QUESTION}}$

1. Find 'x' in the following figures ?

$\large\underline{\underline\mathrm\orange{SOLUTION}}$

In the first figure.

GIVEN

$\angle BAC =56°$

$\angle ACD =123°$

$\angle ABC =x$

Using exterior angle property of triangle.

$\angle ACD =\angle BAC+\angle ABC$

$\bold{\implies 123° = 56° +x}$

$\bold{\implies 123°-56° = x}$

$\bold{\implies x = 67°}$

In the second figure.

GIVEN

$\angle BAC =25°$

$\angle ACB =30°$

$\angle ABC =x$

Using angle sum property of triangle

$\angle ABC +\angle BAC +\angle ACB=180°$

$\bold{\implies x +25+30=180°}$

$\bold{\implies x +55=180°}$

$\bold{\implies x =180°-55}$

$\bold{\implies x =125°}$

Answer 2

i) ∆ABC, angleBAC = 56°

angleACD = 123°

therefore,

angleACB = 180°-angleACD (exterior angle of a triangle) = 180°-123° = 57°

angleBAC + angleACB + angleABC = 180° ( angle sum property of triangle)

= 56°+ 57°+ angleABC = 180°

= 113° + angleABC = 180°

= angleABC = 180°-113° = 67°

Therefore, angleABC= 67°

ii) angleABC+ angleBCA + angleCAB = 180° (angle sum property of triangle)

= x+ 30°+25° =180°

= x + 55° = 180°

= x = 180°-55°

= x = 125°