Question-223042

Question Number 223042 by Tawa11 last updated on 13/Jul/25

Commented by Tawa11 last updated on 13/Jul/25

$$\mathrm{Is}\:\:\mathrm{x}\:\:=\:\:\mathrm{5}.\mathrm{139}°\:\:\mathrm{or}\:\:\mathrm{x}\:\:=\:\:\mathrm{80}° \\ $$

Commented by mr W last updated on 13/Jul/25

i guess 5.139° is wrong. but maybe x=80° is also wrong. why not tell how you got?

$${i}\:{guess}\:\mathrm{5}.\mathrm{139}°\:{is}\:{wrong}.\:{but}\:{maybe} \\ $$$${x}=\mathrm{80}°\:{is}\:{also}\:{wrong}.\:{why}\:{not}\:{tell} \\ $$$${how}\:{you}\:{got}? \\ $$

Commented by Tawa11 last updated on 13/Jul/25

One of your answer was 5.139° Drawn to scale gave 80°. So, I want to know the correct answer. I was checking your post. I saw the question. Someone also ask the same question, so, I recall I have seen it here

$$\mathrm{One}\:\mathrm{of}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{was}\:\:\mathrm{5}.\mathrm{139}° \\ $$$$\mathrm{Drawn}\:\mathrm{to}\:\mathrm{scale}\:\mathrm{gave}\:\:\mathrm{80}°. \\ $$$$\mathrm{So},\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}. \\ $$$$\mathrm{I}\:\mathrm{was}\:\mathrm{checking}\:\mathrm{your}\:\mathrm{post}.\:\mathrm{I}\:\mathrm{saw}\:\mathrm{the}\:\mathrm{question}. \\ $$$$\mathrm{Someone}\:\mathrm{also}\:\mathrm{ask}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question},\:\mathrm{so},\:\mathrm{I} \\ $$$$\mathrm{recall}\:\mathrm{I}\:\mathrm{have}\:\mathrm{seen}\:\mathrm{it}\:\mathrm{here} \\ $$

Commented by Tawa11 last updated on 13/Jul/25

$$\mathrm{Q112076} \\ $$

Commented by mr W last updated on 13/Jul/25

$${what}\:{did}\:{you}\:{get}?\:{just}\:{to}\:{see}\:{what} \\ $$$${others}\:{get},\:{is}\:{not}\:“{learning}''. \\ $$

Commented by fantastic last updated on 13/Jul/25

Commented by Tawa11 last updated on 13/Jul/25

$$\mathrm{I}\:\mathrm{draw}\:\mathrm{to}\:\mathrm{scale}\:\mathrm{and}\:\mathrm{got}\:\:\mathrm{80}° \\ $$

Commented by fantastic last updated on 13/Jul/25

$${I}\:{will}\:{advise}\:{you}\:{to}\:{try}\:{to}\:{solve}\:{it} \\ $$$${in}\:{your}\:{own}.{I}\:{just}\:{saw}\:{the}\:{q}.\:{Im}\: \\ $$$${trying}\:{to}\:{solve} \\ $$$$ \\ $$

Commented by mr W last updated on 13/Jul/25

when you got 80° by drawing to scale, then you know that 5.139° is not right. when you are still not sure, then try to solve it to comfirm that you are right (or not).

$${when}\:{you}\:{got}\:\mathrm{80}°\:{by}\:{drawing}\:{to}\:{scale}, \\ $$$${then}\:{you}\:{know}\:{that}\:\mathrm{5}.\mathrm{139}°\:{is}\:{not} \\ $$$${right}.\:{when}\:{you}\:{are}\:{still}\:{not}\:{sure}, \\ $$$${then}\:{try}\:{to}\:{solve}\:{it}\:{to}\:{comfirm}\:{that} \\ $$$${you}\:{are}\:{right}\:\left({or}\:{not}\right).\: \\ $$

Commented by Tawa11 last updated on 13/Jul/25

I know sir. I don′t want to conclude on your result sir. Because you may reason another thing. You may actually have another reason. That is why I asked.

$$\mathrm{I}\:\mathrm{know}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{conclude}\:\mathrm{on}\:\mathrm{your}\:\mathrm{result}\:\mathrm{sir}. \\ $$$$\mathrm{Because}\:\mathrm{you}\:\mathrm{may}\:\mathrm{reason}\:\mathrm{another}\:\mathrm{thing}. \\ $$$$\mathrm{You}\:\mathrm{may}\:\mathrm{actually}\:\mathrm{have}\:\mathrm{another}\:\mathrm{reason}. \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{asked}. \\ $$

Commented by mr W last updated on 13/Jul/25

$${i}\:{also}\:{make}\:{mistakes}. \\ $$

Commented by Tawa11 last updated on 13/Jul/25

Noted sir. Thanks for everytime attention.

$$\mathrm{Noted}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{everytime}\:\mathrm{attention}. \\ $$

Answered by fantastic last updated on 13/Jul/25

Commented by fantastic last updated on 13/Jul/25

∡AOC=2×∡ABC⇒60^0 AO=OC and ∡AOC=60^0 So △AOC is eq. triangle So AC=AO=OC ∡ADC=20^0 +20^0 +30^0 =70^0 ∡CAD=180^0 −(70^0 +60^0 −20^0 )=180^0 −110^0 =70^0 So CD=AC but AC=CO So CD=CO ∴∡COD+∡ODC+20^0 =180^0 2x+20^0 =180^0 ⇒ determinant (((x=80^0 )))

$$\measuredangle{AOC}=\mathrm{2}×\measuredangle{ABC}\Rightarrow\mathrm{60}^{\mathrm{0}} \\ $$$${AO}={OC}\:{and}\:\measuredangle{AOC}=\mathrm{60}^{\mathrm{0}} \\ $$$${So}\:\bigtriangleup{AOC}\:{is}\:{eq}.\:{triangle} \\ $$$${So}\:{AC}={AO}={OC} \\ $$$$\measuredangle{ADC}=\mathrm{20}^{\mathrm{0}} +\mathrm{20}^{\mathrm{0}} +\mathrm{30}^{\mathrm{0}} =\mathrm{70}^{\mathrm{0}} \\ $$$$\measuredangle{CAD}=\mathrm{180}^{\mathrm{0}} −\left(\mathrm{70}^{\mathrm{0}} +\mathrm{60}^{\mathrm{0}} −\mathrm{20}^{\mathrm{0}} \right)=\mathrm{180}^{\mathrm{0}} −\mathrm{110}^{\mathrm{0}} =\mathrm{70}^{\mathrm{0}} \\ $$$${So}\:{CD}={AC} \\ $$$${but}\:{AC}={CO} \\ $$$${So}\:{CD}={CO}\: \\ $$$$\therefore\measuredangle{COD}+\measuredangle{ODC}+\mathrm{20}^{\mathrm{0}} =\mathrm{180}^{\mathrm{0}} \\ $$$$\mathrm{2}{x}+\mathrm{20}^{\mathrm{0}} =\mathrm{180}^{\mathrm{0}} \\ $$$$\Rightarrow\begin{array}{|c|}{{x}=\mathrm{80}^{\mathrm{0}} }\\\hline\end{array} \\ $$

Commented by fantastic last updated on 13/Jul/25