Question
Answer:
:
Given: BEST is a rhombus, so,
TS∥BE
and BS is a transversal.
∴∠SBE=∠TSB=40∘ [alternate interior anges]
Also,∠y=90∘ [diagonals of a rhombus bisect at 90∘]
In ΔTSO,
∠STO+∠TSO+∠SOT=180∘ [Angle sum property of a triangle]
x+40∘+90∘=180∘⇒x=180∘−90∘−40∘=50∘
∴y−x=90∘−50∘=40∘
Was this answer helpful ?
:
Given: BEST is a rhombus, so,
TS∥BE
and BS is a transversal.
∴∠SBE=∠TSB=40∘ [alternate interior anges]
Also,∠y=90∘ [diagonals of a rhombus bisect at 90∘]
In ΔTSO,
∠STO+∠TSO+∠SOT=180∘ [Angle sum property of a triangle]
x+40∘+90∘=180∘⇒x=180∘−90∘−40∘=50∘
∴y−x=90∘−50∘=40∘
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution