Question

answer for the sixteenth question please

Answer 1

By linear pair axiom we have,

∠DEP=180-∠5

∠EDP=180-∠4

In △PDE, sum of all angles will be 180.

⇒∠P+180-∠5+180-∠4=180

⇒∠P=∠5+∠4-180

Similarly∠Q=∠5+∠1-180

∠R=∠1+∠2-180

∠S=∠2+∠3-180

∠T=∠3+∠4-180

Adding above 5 equations, we get,

∠P+∠Q+∠R+∠S+∠T=2∠1+∠2+∠3+∠4+∠5-900

In pentagon ABCDE,

∠1+∠2+∠3+∠4+∠5=540

⇒∠P+∠Q+∠R+∠S+∠T=2540-900=1080-900

⇒∠P+∠Q+∠R+∠S+∠T=180

= 2 right angle