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pritibhamra94pb4for:
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AEF = EGR
angle RGE=60 DEGREE
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RGE = BGR
BGR=60
x=20
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construct a line passing through B parallel to RS||PQ.... then the obtuse angle formed by line segment GB with the constructed line will be equal to 120° (Corresponding Angles on transverse of parallel lines).
The acute angle formed by HB with the constructed line will be 100° (Corresponding Angles)
x = 120-100 = 20°
The acute angle formed by HB with the constructed line will be 100° (Corresponding Angles)
x = 120-100 = 20°
Answered by
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Dear friend,
Here's your answer...
As PQ//RS,
/_AEF = /_EGH = 120° (Corresponding angles)
/_EGH + /_BGH = 180° ( linear pair)
120° + /_BGH = 180°
/_BGH = 60°
/_BHG + /_BHS = 180° (Linear pair)
/_BHG + 100°= 180°
/_BHG = 80°
In Δ BHG,
/_BGH + /_BHG + /_GBH = 180° ( Angle sum property of a triangle)
60° + 80° + x° = 180°
140° + x° = 180°
x = 180 - 140
x = 40°
Hope it helps you !!
Here's your answer...
As PQ//RS,
/_AEF = /_EGH = 120° (Corresponding angles)
/_EGH + /_BGH = 180° ( linear pair)
120° + /_BGH = 180°
/_BGH = 60°
/_BHG + /_BHS = 180° (Linear pair)
/_BHG + 100°= 180°
/_BHG = 80°
In Δ BHG,
/_BGH + /_BHG + /_GBH = 180° ( Angle sum property of a triangle)
60° + 80° + x° = 180°
140° + x° = 180°
x = 180 - 140
x = 40°
Hope it helps you !!
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