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secondary 3 | E Maths
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need help asap. the line forming the isos triangle is drawn by me
Look at the four-sided trapezium-looking figure involving the lower angle, 130 degrees, 138 degrees and one more angle.
The lower angle is directly opposite the 138 degrees in this quadrilateral, so the value of the lower angle
= 180 - 138
= 42 degrees
This property is also known as "angles in opposite segments add to 180 degrees" or "opposite angles of a cyclic quadrilateral add to 180 degrees".
The upper angle must have been 53 degrees.
Value of x
= 180 - 53 - 53
= 74 degrees
The upper angle of y is also 53 degrees.
Lower angle of y
= 180 - 130 (similar in idea to the other lower angle)
= 50 degrees
Value of y
= 50 + 53
= 103 degrees
Or, we can use sum of interior angles of this 5-sided figure
= 180 x (5 - 2)
= 540 degrees
and the value of y
= 540 - 138 - 130 - 95 - 74
= 103 degrees
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