JAMB - Mathematics (1990 - No. 40)
40o
50o
80o
100o
स्पष्टीकरण
Since PS = QS = RS
S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius
QPR \(\pm\) \(\frac{100^o}{2}\) = 50o
S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius
QPR \(\pm\) \(\frac{100^o}{2}\) = 50o