in the given figure AS||BT;<4=<5 SB bisects <AST.find the measure of <1
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AS∥BT [ Given ]
∴ ∠2=∠5 [ Alternate angles are equal ]
⇒ ∠2=∠3
⇒ Let, ∠2=∠3=∠5=x
⇒ In △BST, ∠RSB is an exterior angle.
∴ ∠RSB=∠SBT+∠BTS [ By exterior angle property ]
∴ ∠1+∠2=∠5+∠4
∴ ∠1+x=x+x
∴ ∠1+x=2x
∴ ∠1=x
⇒ ∠1+∠2+∠3=180
o
[ Since, ∠1,∠2 and ∠3 forms a linear pair ]
⇒ x+x+x=180
o
⇒ 3x=180
o
∴ x=
3
180
o
=60
o
∴ Measure of ∠1 is 60
o
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