4) Four bells beepat intervals of 40,45,60 and 75
seconds Nespestivaly if they beep
simultaneously at 5:30 am at what t

2 thoughts on “4) Four bells beepat intervals of 40,45,60 and 75 <br />seconds Nespestivaly if they beep<br />simultaneously at 5:30 am at what t”

1. Answer:

   4 min 40 s

   The time after which they will ring together will be LCM of 4,5,7,8,10 which is 280sec that is equal to 4min40sec

   So correct answer will be option C

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2. Answer:

   [tex]\boxed{\bold{\pink{cos \alpha = \dfrac{9}{41} \: and \: cosec \alpha = \dfrac{41}{40} }}}[/tex]

   Step-by-step explanation:

   [tex]\green{\bold{Given}}\longrightarrow \\ 9 \: cos \alpha + 40 \: sin \alpha = 41[/tex]

   [tex]\red{\bold{To \: find}}\longrightarrow \\ value \: of \: cos \alpha \: and \: cosec \alpha [/tex]

   [tex]\blue{\bold{Concept \: used }}\longrightarrow\\ 1) {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ 2) {sin}^{2} \alpha = 1 – {cos}^{2} \alpha \\ 3) {(x – y)}^{2} = {x}^{2} + {y}^{2} – 2xy \\ 4) \: cosec \alpha = \dfrac{1}{sin \alpha } [/tex]

   [tex]\pink{\bold{Solution}}\longrightarrow \\ 9cos \alpha + 40sin \alpha = 41[/tex]

   [tex] = > 40sin \alpha = 41 – 9cos \alpha [/tex]

   [tex]squaring \: both \: sides \: we \: get[/tex]

   [tex] = > {(40sin \alpha )}^{2} = {(41 – 9cos \alpha )}^{2} [/tex]

   [tex] = > 1600 {sin}^{2} \alpha = {(41)}^{2} + {(9cos \alpha) }^{2} – 2 \: (41) \: (9cos \alpha ) \\ = > 1600(1 – {sin}^{2} \alpha ) = 1681 + 81 {cos}^{2} \alpha – 738cos \alpha [/tex]

   [tex] = > 1600 – 1600 {cos}^{2} \alpha – 81 {cos}^{2} \alpha + 738cos \alpha – 1681 = 0 \\ = > – 1681 {cos}^{2} \alpha + 738cos \alpha + 81 = 0[/tex]

   [tex] = > 1681 {cos}^{2} \alpha – 738cos \alpha + 81 = 0 \\ = > {(41cos \alpha )}^{2} – 2 \: (41cos \alpha ) \: ( \: 9 \: ) + {(9)}^{2} = 0 \\ = > {(41cos \alpha – 9)}^{2} = 0 \\ taking \: square \: root \: of \: both \: sides \\ = > 41cos \alpha – 9 = 0 \\ = >41 cos \alpha = 9 \\ = >\pink{ cos \alpha = \dfrac{9}{41}} \\ now \\ {sin}^{2} \alpha = 1 – {cos}^{2} \alpha \\ = > sin \alpha = \sqrt{1 – {cos}^{2} \alpha } \\ = > sin \alpha = \sqrt{1 – {( \dfrac{9}{41}) }^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 – \dfrac{81}{1681} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ \dfrac{1681 – 81}{81} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ \dfrac{1600}{1681} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{40}{41} [/tex]

   [tex]now [/tex]

   [tex]cosec \alpha = \dfrac{1}{sin \alpha } \\ \\= > cosec \alpha = \dfrac{1}{ \dfrac{ 40}{41} } \\ \\ = >\pink{ cosec \alpha = \dfrac{41}{40}} [/tex]

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