Question

In the given figure, PQ || RS, ZAEF = 95°.
BHS = 110° and ABC=xº. Find the value of x​

Answer 1

Answer:

BHS=110°

BHG=180°-BHS

=180°-110°

=70°

AEF=95°

EGH=AEF [ALTERNATIVE INTERIOR ANGLES]

=95°

BGH=180°EGH

=180°-95°

=85°

A/q, HGB+GHB+GBH=180°[ANGLE SUM PROPERTY]

=> 85°+70°+x°=180°

=> 155°+x°=180°

=> x°=180°-155°

=> x°=25°

Answer 2

Step-by-step explanation:

Given :

From  Fig. 8.129, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°.  

To find :

Then the value of x.

 

Solution :  

∠AEF = ∠AGH = 95° (Corresponding angles)

∠AGH + ∠HGB = 180° (Linear pair)

95° + ∠HGB = 180°

∠HGB = 180° - 95°

∠HGB = 85°

∠BHS + ∠BHG = 180° (Linear pair)

110° + ∠BHG = 180°

∠BHG = 180° - 110°

∠BHG = 70°

In ∆BHG,

∠BHG + ∠HGB + ∠GBH = 180°

[By angle sum property]

70° + 85° + ∠GBH = 180°

155° + ∠GBH = 180°

∠GBH = 180° - 155°

∠GBH = 25°

Now,

∠ABC = ∠GBH = 25°

∠ABC = x° = 25°.

Hence, the value of x is 25°.