In the figure PT is a tangent to the circle with centre at O. If PQT = 30o, find the value of PTO
O\(\hat{P}\)T = 90º (The radius is perpendicular to the tangent at the point of contact.)
Sum of angles in \(\triangle\) PQT = 30 + ( 90 + x ) + (x + 2x) = 180
= 120 + 4x = 180
4x = 180 - 120 = 60º
x = \(\frac{60}{4}\) = 15°
Therefore, P\(\hat{T}\)O = 2x = 2 x 15 = 30º
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