Math, asked by UsarshiDas9433, 2 months ago

4) Four bells beepat intervals of 40,45,60 and 75
seconds Nespestivaly if they beep
simultaneously at 5:30 am at what time
they will beep togethar again?​

Answers

Answered by singhreema20577
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

Answered by Anonymous
49

Answer:

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

Step-by-step explanation:

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

\red{\bold{To \: find}}\longrightarrow  \\ value \: of \: cos \alpha  \: and \: cosec \alpha

\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 }

\pink{\bold{Solution}}\longrightarrow \\ 9cos \alpha  + 40sin \alpha  = 41

 =  > 40sin \alpha  = 41 - 9cos \alpha

squaring \: both \: sides \: we \: get

 =  >  {(40sin \alpha )}^{2}  =  {(41 - 9cos \alpha )}^{2}

 =  > 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

 =  > 1600 - 1600 {cos}^{2} \alpha - 81 {cos}^{2}   \alpha  + 738cos \alpha  - 1681 = 0 \\  =  >  - 1681 {cos}^{2}  \alpha  + 738cos \alpha  + 81 = 0

 =  > 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}

now

cosec \alpha  =  \dfrac{1}{sin \alpha }  \\  \\=  > cosec \alpha  =  \dfrac{1}{ \dfrac{ 40}{41} }  \\ \\ =  >\pink{ cosec \alpha  =  \dfrac{41}{40}}

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