### Find radius of the circle given a triangle whose base is tangential to the circle

Question Sample Titled 'Find radius of the circle given a triangle whose base is tangential to the circle'

In the figure, ${A}$ , ${B}$ and ${C}$ are points lying on the circle. ${T}{A}$ is the tangent to the circle at ${A}$. The straight line ${C}{B}{T}$ is perpendicular to ${T}{A}$. If ${B}{C}={7}$ $\text{cm}$, find the radius of the circle correct to the nearest ${0.1}$ $\text{cm}$.

${A}$${C}$${T}$${B}$${14}^{\circ}$
A
${4}$ $\text{cm}$
B
${7.5}$ $\text{cm}$
C
${3.4}$ $\text{cm}$
D
${5.5}$ $\text{cm}$

 Method ${1}$: Constucting a diameter Join ${A}{B}$ .  $\angle{B}{A}{T}$ $={14}^{\circ}$ ∠ in alt. segment Consider $\triangle{T}{A}{C}$ , $\angle{C}{A}{T}$ $={180}^{\circ}-{90}^{\circ}-{14}^{\circ}={76}^{\circ}$ ∠ sum of △ $\angle{B}{A}{C}$ $={76}^{\circ}-{14}^{\circ}={62}^{\circ}$ Let ${D}$ be a point on circle such that ${C}{D}$ is a diameter of length ${2}{r}$ . $\angle{B}{D}{C}$ $={62}^{\circ}$ ∠s in the same segment $\angle{C}{B}{D}$ $={90}^{\circ}$ ∠ in semi-circle Consider $\triangle{B}{C}{D}$ , $\dfrac{{{2}{r}}}{{{\sin{{90}}}^{\circ}}}$ $=\dfrac{{7}}{{{\sin{{62}}}^{\circ}}}$ sine formula ${r}$ $=\dfrac{{7}}{{{2}{\sin{{62}}}^{\circ}}}$  $={3.9639951774116375}$  $={4}$ $\text{cm }\$(cor. to the nearest ${0.1}$ $\text{cm}$) ∴   The radius of the circle is ${4}$ $\text{cm}$ .

${A}$${C}$${T}$${B}$${14}^{\circ}$${O}$${D}$${14}^{\circ}$${62}^{\circ}$${62}^{\circ}$${r}$${r}$Method ${1}$

 Method ${2}$: Bisecting the chord ${A}{C}$ Join ${A}{B}$ .  $\angle{B}{A}{T}$ $={14}^{\circ}$ ∠ in alt. segment Consider $\triangle{T}{A}{C}$ , $\angle{C}{A}{T}$ $={180}^{\circ}-{90}^{\circ}-{14}^{\circ}={76}^{\circ}$ ∠ sum of △ Consider $\triangle{A}{B}{C}$ , $\angle{B}{A}{C}$ $={76}^{\circ}-{14}^{\circ}={62}^{\circ}$ $\angle{A}{B}{C}$ $={180}^{\circ}-{14}^{\circ}-{62}^{\circ}={104}^{\circ}$ ∠ sum of △ $\dfrac{{{A}{C}}}{{{\sin{{104}}}^{\circ}}}$ $=\dfrac{{7}}{{{\sin{{62}}}^{\circ}}}$ sine formula ${A}{C}$ $=\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}$ $\text{cm}$ Let ${O}$ be the center of the circle. Let ${r}$ be the radius. Let ${M}$ be a point on chord ${A}{C}$ such that ${O}{M}\bot{A}{C}$ . Join ${O}{M}$ . ${A}{M}$ $=\dfrac{{{A}{C}}}{{2}}=\dfrac{{1}}{{2}}{\left(\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}\right)}=\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{2}{\sin{{62}}}^{\circ}}}$ line from centre ⊥ chord bisects chord $\angle{O}{A}{C}$ $={14}^{\circ}$ alt. ∠s Consider $\triangle{O}{A}{M}$ , ${{\cos{{14}}}^{\circ}}$ $=\dfrac{{\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{2}{\sin{{62}}}^{\circ}}}}}{{r}}$ ${r}$ $=\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{2}{\sin{{62}}}^{\circ}}}\cdot\dfrac{{1}}{{{\cos{{14}}}^{\circ}}}$  $={3.9639951774116375}$  $={4}$ $\text{cm }\$(cor. to the nearest ${0.1}$ $\text{cm}$) ∴   The radius of the circle is ${4}$ $\text{cm}$ .

${A}$${C}$${T}$${B}$${14}^{\circ}$${O}$${14}^{\circ}$${r}$${104}^{\circ}$${62}^{\circ}$${7}$ cm${M}$${14}^{\circ}$$\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{2}{\sin{{62}}}^{\circ}}}$cmMethod ${2}$

 Method ${3}$: Using cosine formula Join ${A}{B}$ .  $\angle{B}{A}{T}$ $={14}^{\circ}$ ∠ in alt. segment Consider $\triangle{T}{A}{C}$ , $\angle{C}{A}{T}$ $={180}^{\circ}-{90}^{\circ}-{14}^{\circ}={76}^{\circ}$ ∠ sum of △ Consider $\triangle{A}{B}{C}$ , $\angle{B}{A}{C}$ $={76}^{\circ}-{14}^{\circ}={62}^{\circ}$ $\angle{A}{B}{C}$ $={180}^{\circ}-{14}^{\circ}-{62}^{\circ}={104}^{\circ}$ ∠ sum of △ $\dfrac{{{A}{C}}}{{{\sin{{104}}}^{\circ}}}$ $=\dfrac{{7}}{{{\sin{{62}}}^{\circ}}}$ sine formula ${A}{C}$ $=\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}$ $\text{cm}$ Let ${O}$ be the center of the circle. Let ${r}$ be the radius. Join ${O}{A}$ and ${O}{C}$ . reflex $\angle{A}{O}{C}$ $={2}{\left({104}^{\circ}\right)}={208}^{\circ}$ ∠ at centre twice ∠ at ⊙ce $\angle{A}{O}{C}$ $={360}^{\circ}-{208}^{\circ}={152}^{\circ}$ ∠s at a pt. Consider $\triangle{O}{A}{C}$ , ${r}^{{2}}+{r}^{{2}}-{2}{\left({r}\right)}{\left({r}\right)}{\left({\cos{{152}}}^{\circ}\right)}$ $={\left(\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}\right)}^{{2}}$ cosine formula ${2}{r}^{{2}}-{2}{r}^{{2}}{{\cos{{152}}}^{\circ}}$ $={\left(\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}\right)}^{{2}}$ ${r}^{{2}}{\left({2}-{2}{\cos{{152}}}^{\circ}\right)}$ $={\left(\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}\right)}^{{2}}$ ${r}$ $=\sqrt{{\dfrac{{{\left(\dfrac{{{7}{\sin{{104}}}^{\circ}}}{{{\sin{{62}}}^{\circ}}}\right)}^{{2}}}}{{{2}-{2}{\cos{{152}}}^{\circ}}}}}$  $={3.9639951774116375}$  $={4}$ $\text{cm }\$(cor. to the nearest ${0.1}$ $\text{cm}$) ∴   The radius of the circle is ${4}$ $\text{cm}$ .

${A}$${C}$${T}$${B}$${14}^{\circ}$${O}$${14}^{\circ}$${r}$${r}$${104}^{\circ}$${62}^{\circ}$${7}$ cm${208}^{\circ}$${152}^{\circ}$Method ${3}$

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