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Question Number 100908 by bemath last updated on 29/Jun/20

$$\mathrm{what}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{angle}$$$$\mathrm{formed}\mathrm{by}\mathrm{a}\mathrm{long}\mathrm{needle}\mathrm{and}$$$$\mathrm{short}\mathrm{needle}\mathrm{on}\mathrm{analog}\mathrm{clock}$$$$\mathrm{that}\mathrm{shows}\mathrm{at}15.50?$$$$\left(\mathrm{A}\right){175}^{\mathrm{o}}\left(\mathrm{B}\right){174}^{\mathrm{o}}\left(\mathrm{C}\right){173}^{\mathrm{o}}$$$$\left(\mathrm{D}\right){172}^{\mathrm{o}}\left(\mathrm{E}\right){170}^{\mathrm{o}}$$

Commented by behi83417@gmail.com last updated on 29/Jun/20

$$\theta =\mid \frac{11}{2}.\mathrm{m}-30.\mathrm{h}\mid [\mathrm{m}=\mathrm{minute},\mathrm{h}=\mathrm{houre}]$$$$\mathit{\theta }=\mid \frac{11}{2}\times 50-30\times 15\mid =\mid -175\mid ={175}^{\bullet }◼$$

Commented by bemath last updated on 29/Jun/20

$$\mathrm{yes}\mathrm{sir}\mathrm{correct}\mathrm{answer}{175}^{\mathrm{o}}$$

Answered by 1549442205 last updated on 29/Jun/20

$$\mathrm{At}15{\mathrm{o}}^{\prime }\mathrm{clock}\mathrm{the}\mathrm{long}{\mathrm{needle}}^{\prime }\mathrm{end}\mathrm{and}\mathrm{short}$$$${\mathrm{needle}}^{\prime }\mathrm{s}\mathrm{end}\mathrm{are}\mathrm{spaced}\mathrm{a}\mathrm{quarter}\mathrm{distance}$$$$\mathrm{of}\mathrm{circle}\mathrm{away}\mathrm{from}\mathrm{each}\mathrm{other}.$$$$\mathrm{The}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{long}{\mathrm{needle}}^{\prime }\mathrm{s}\mathrm{end}$$$$\mathrm{equal}\frac{1}{60}\left(\mathrm{circle}\mathrm{per}\mathrm{a}\mathrm{minute}\right)\mathrm{and}\mathrm{of}$$$$\mathrm{short}\mathrm{needle}\mathrm{equal}\frac{1}{12\times 60}\left(\mathrm{circle}\mathrm{per}\mathrm{a}\mathrm{minute}\right)$$$$\mathrm{Hence}\mathrm{after}50\mathrm{minutes}\mathrm{the}\mathrm{long}{\mathrm{needle}}^{\prime }\mathrm{s}\mathrm{end}$$$$\mathrm{went}\mathrm{a}\mathrm{distance}\mathrm{equal}\frac{1}{60}\times 50=\frac{5}{6}\left(\mathrm{circle}\right)$$$$\mathrm{while}\mathrm{the}\mathrm{short}{\mathrm{needle}}^{\prime }\mathrm{s}\mathrm{end}\mathrm{went}\mathrm{a}\mathrm{distance}$$$$\mathrm{equal}50\times \frac{1}{720}=\frac{5}{72}\left(\mathrm{circle}\right).\mathrm{Therefore},$$$$\mathrm{at}15.50\mathrm{the}\mathrm{arc}\mathrm{create}\mathrm{by}\mathrm{ends}\mathrm{of}\mathrm{both}$$$$\mathrm{the}\mathrm{needles}\mathrm{has}\mathrm{the}\mathrm{length}\mathrm{equal}$$$$\frac{5}{6}-(\frac{1}{4}+\frac{5}{72})=\frac{37}{72}\left(\mathrm{circle}\right)\mathrm{which}\mathrm{means}$$$$\mathrm{that}\mathrm{it}\mathrm{create}\mathrm{an}\mathrm{angle}\mathrm{has}\mathrm{the}\mathrm{value}$$$$\mathbf{equal}\frac{37}{72}\times 360°=185°$$$$\mathbf{Please},\mathbf{sir}\mathbf{check}\mathbf{to}\mathbf{help}\mathbf{me}\mathbf{is}\mathbf{it}\mathbf{correct}?$$

Commented by bemath last updated on 29/Jun/20

$$\mathrm{means}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{correct}\mathrm{answer}$$$$\mathrm{option}?$$

Commented by EuclidKaBaap last updated on 29/Jun/20

$$hahahaha..itssimplethingwe$$$$oftenoversee..itsareflexangleand$$$$theyhaventspecifiedwhichside..$$$$so360-185=175willbetherequired$$$$answeraskedinthequestion$$

Commented by 1549442205 last updated on 29/Jun/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sir},\mathrm{a}\mathrm{simple}\mathrm{thing}\mathrm{that}\mathrm{I}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{out}!\mathrm{The}\mathrm{angle}\mathrm{is}\mathrm{created}\mathrm{by}$$$$\mathrm{long}\mathrm{hand}\mathrm{and}\mathrm{the}\mathrm{short}\mathrm{hand}\left(\mathrm{clockwise}\right)$$$$\mathrm{equal}360°-185°=175°,\mathrm{so}\mathrm{choose}\mathrm{answer}$$$$\mathbf{A}:175°$$

Answered by ajfour last updated on 29/Jun/20

$$At15:00angleis90°.$$$$360°ofminutehand\equiv 30°hourhand$$$$300°ofminutehand\equiv {\left(\frac{300}{12}\right)}^{°}hourhand$$$$shorteranglebetweenthemat$$$$15:50is$$$$=60°+90°+{\left(\frac{300}{12}\right)}^{°}$$$$=175°.$$

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