WAEC - Mathematics (2015 - No. 18)
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and <TOS = 76\(^o\). Using the diagram, calculate the value of angle PTR.
73o
67o
57o
37o
Explanation
In the diagram given, < PRT = 3\(^o\) (Change in same segment)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)
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