Question

(a) If (2x +20)° and (3x - 40)º make a straight angle then find the value of x.
(b) If (x + 20°) and (3x - 40)° are the supplementary angles to each other then find these angle​

Answer 1

hope this will be helpful

Answer 2

a)

Given:

• (2x + 20)° and (3x - 40)° make an straight angle

Find:

• What will be the value of x = ?

Solution:

We, know that measure of an straight angle is 180°

So,

$\sf \mapsto 2x + {20}^{ \circ} + 3x - {40}^{ \circ} = {180}^{ \circ}$

$\sf \mapsto 2x + 3x+ {20}^{ \circ} - {40}^{ \circ} = {180}^{ \circ}$

$\sf \mapsto 5x - {20}^{ \circ} = {180}^{ \circ}$

$\sf \mapsto 5x = {180}^{ \circ} + {20}^{ \circ}$

$\sf \mapsto 5x = {200}^{ \circ}$

$\sf \mapsto x = \frac{{200}^{ \circ} }{5}$

$\sf \mapsto x = {\frac{200}{5}}^{ \circ}$

$\sf \mapsto x ={40}^{ \circ}$

Hence, value of x will be 40°

__________________

b)

Given:

• (x + 20)° and (3x - 40)° are the supplementary angles

Find:

• Find the angles

Solution:

We, know that the supplementary angles are angles which gives 180° when added.

So,

$\sf \hookrightarrow {x + 20}^{ \circ} + {3x - 40}^{ \circ} = {180}^{ \circ}$

$\sf \hookrightarrow x +3x + {20}^{ \circ} - { 40}^{ \circ} = {180}^{ \circ}$

$\sf \hookrightarrow 4x - { 20}^{ \circ} = {180}^{ \circ}$

$\sf \hookrightarrow 4x = {180}^{ \circ} + { 20}^{ \circ}$

$\sf \hookrightarrow 4x = {200}^{ \circ}$

$\sf \hookrightarrow x = \frac{{200}^{ \circ}}{4}$

$\sf \hookrightarrow x = {50}^{ \circ}$

So, x = 50°

Hence, value of f1st angle = x + 20° = 50 + 20° = 80°

Value of 2nd angle = 3x - 40° = 3(50) - 40° = 150 - 40° = 110°