Question

In the given figure, points B and C lie on tangent to the circle drawn at point A. Chord AD$\cong$ chord ED. If m(arc EPF) = $\frac{1}{2}$m(arc AQD) and m(arc DRE) = 84$°$ then determine
(i) $\angle$DAC
(ii) $\angle$FDA
(iii) $\angle$FED
(iv) $\angle$BAF

Answer 1

Construction: Join OD, OE ,OA,OF, FD and AF.

(i)
$$\begin{gathered} \mathrm{In}△\mathrm{EOD}\mathrm{and}△\mathrm{AOD},\mathrm{we}\mathrm{have}: \\ \mathrm{AO}=\mathrm{EO}(\mathrm{Radii}\mathrm{of}\mathrm{the}\mathrm{circle}) \\ \mathrm{OD}=\mathrm{OD}\left(\mathrm{Common}\right) \\ \mathrm{ED}=\mathrm{AD}\left(\mathrm{Given}\right) \\ \therefore △\mathrm{EOD}\cong △\mathrm{AOD}(\mathrm{By}\mathrm{SSS}\mathrm{congruency}) \\ \mathrm{i}.\mathrm{e}.,\angle \mathrm{DEO}=\angle \mathrm{DAO}\left(\mathrm{CPCT}\right) \\ \mathrm{Also},\angle \mathrm{DOE}=\angle \mathrm{AOD}=84°(\mathrm{CPCT},\mathrm{m}(\mathrm{arcDRE})=84°) \\ \mathrm{Now},\mathrm{in}△\mathrm{EOD},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{EOD}+\angle \mathrm{ODE}+\angle \mathrm{OED}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{OED}=180°-84° \\ \Rightarrow \angle \mathrm{OED}=48° \\ \mathrm{Hence},\angle \mathrm{OED}=\angle \mathrm{DAO}=48° \\ \angle \mathrm{EAC}=90°(\mathrm{Radius}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{tangent}) \\ \angle \mathrm{DAC}=\angle \mathrm{EAC}-\angle \mathrm{DAO}=90°-48°=42° \\ \angle \mathrm{EOF}=\frac{1}{2}\angle \mathrm{AOD}=42° \end{gathered}$$


(ii)

$$\begin{gathered} \angle \mathrm{EOF}+\angle \mathrm{EOD}+\angle \mathrm{DOA}+\angle \mathrm{AOF}=360°(\mathrm{Complete}\mathrm{angle}) \\ \Rightarrow \angle \mathrm{AOF}=360°-\angle \mathrm{EOF}-\angle \mathrm{EOD}-\angle \mathrm{DOA} \\ \Rightarrow \angle \mathrm{AOF}=360°-42°-84°-84°=150° \end{gathered}$$

We know that the angle subtended by an arc at the centre is twice the angle subtended at any part of the circle.
i.e., $\angle \mathrm{FDA}=\frac{1}{2}\angle \mathrm{AFO}=75°$

(iii)

$$\begin{gathered} \mathrm{In}△\mathrm{EOF},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{EOF}+\angle \mathrm{FEO}+\angle \mathrm{EFO}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{FEO}=180°-42° \\ \Rightarrow \angle \mathrm{FEO}=69° \\ \mathrm{So},\angle \mathrm{FED}=\angle \mathrm{OED}+\angle \mathrm{FEO}=48°+69°=117° \end{gathered}$$


(iv)

$$\begin{gathered} \mathrm{In}△\mathrm{AOF},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{AOF}+\angle \mathrm{FAO}+\angle \mathrm{OFA}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{OAF}=180°-150° \\ \Rightarrow \angle \mathrm{OED}=15° \\ \mathrm{So},\angle \mathrm{OAB}=90°(\mathrm{Radius}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{tangent}) \\ \therefore \angle \mathrm{FAB}=\angle \mathrm{OAB}-\angle \mathrm{OAF}=90°-15°=75° \end{gathered}$$