• # question_answer A tangent CQ touches a circle with centre O at P. Diameter AB is produced to meet the tangent at C. If $\angle \text{ACP}=\text{a}{}^\circ$and$\angle \text{BPC}=\text{b}{}^\circ$, find the relation connecting a and b. A)  $\text{a}{}^\circ +\text{b}{}^\circ =\text{18}0{}^\circ$              B)  $\text{a}{}^\circ +\text{2b}{}^\circ =\text{9}0{}^\circ$ C)  $\text{a}{}^\circ -\text{b}{}^\circ =\text{6}0{}^\circ$   D)         $\text{2a}{}^\circ +\text{b}{}^\circ =\text{1}00{}^\circ$

Given $x=\frac{5040\times 12}{144}=420$and $219\times =657$. Also, $3\times =219$ (Angles in an isosceles triangle OAP, angle in deternate segment.) $\text{1}0\text{32 }=\text{ 4}0\text{8 }\times \text{ 2 }+\text{ 216}$ $\text{4}0\text{8 }=\text{ 216 }\times \text{ 1 }+\text{ 192}$ $\text{216 }=\text{ 192 }\times \text{ 1 }+\text{ 24}$ In $\text{192 }=\text{ 24 }\times \text{ 8 }+\text{ }0$ ${{6}^{x}}$${{6}^{x}}={{(2\times 3)}^{x}}={{2}^{x}}\times {{3}^{x}}$