Question

If cosec 0=13\5, find tan 0,and cos 0​

Answer 1

$\: \: \: \sf{ \cosec \: a = \frac{13}{5} }$

$\: \: \: \: \: \: \: \: \: \: \sf{ \frac{h}{p} = \frac{13}{5} }$

$\sf{let \: hypotenuse \: be \: 13k \: and \: perpendicular \: be \: 5k}$

$\: \: \sf{by \: pythagoras \: theorm}$

$\: \: \: \: \: \: \sf{h {}^{2} = {p}^{2} + {b}^{2} }$

$\: \: \sf{(13k) {}^{2} = {(5k)}^{2} + {b}^{2} }$

$\: \: \: \sf{169 {k}^{2} = 25 {k}^{2} + {b}^{2} }$

$\: \: \: \: \sf {{b}^{2} =1 69 {k}^{2} - 25 {k}^{2} }$

$\: \: \: \: \: \: \: \sf{ {b}^{2} = 144k {}^{2} }$

$\: \: \: \: \: \: \: \boxed{ \sf{b = 12k}}$

$\boxed{ \sf \pink{ \tan \: a = \frac{p}{b} = \frac{5k}{12k} = {\frac{5}{12} }}}$

$\boxed{ \sf \red{ \cos \: a = \frac{b}{h} = \frac{12k}{13k} = \frac{12}{13}} }$

$\sf\orange{please\:mark\:as\:brainliest............}$

Answer 2

Step-by-step explanation:

\begin{gathered} \: \: \: \sf{ \cosec \: a = \frac{13}{5} } \\ \end{gathered}

coseca=

5

13

\begin{gathered} \: \: \: \: \: \: \: \: \: \: \sf{ \frac{h}{p} = \frac{13}{5} } \\ \end{gathered}

p

h

=

5

13

\sf{let \: hypotenuse \: be \: 13k \: and \: perpendicular \: be \: 5k}lethypotenusebe13kandperpendicularbe5k

\: \: \sf{by \: pythagoras \: theorm}bypythagorastheorm

\: \: \: \: \: \: \sf{h {}^{2} = {p}^{2} + {b}^{2} }h

2

=p

2

+b

2

\: \: \sf{(13k) {}^{2} = {(5k)}^{2} + {b}^{2} }(13k)

2

=(5k)

2

+b

2

\: \: \: \sf{169 {k}^{2} = 25 {k}^{2} + {b}^{2} }169k

2

=25k

2

+b

2

\: \: \: \: \sf {{b}^{2} =1 69 {k}^{2} - 25 {k}^{2} }b

2

=169k

2

−25k

2

\: \: \: \: \: \: \: \sf{ {b}^{2} = 144k {}^{2} }b

2

=144k

2

\: \: \: \: \: \: \: \boxed{ \sf{b = 12k}}

b=12k

\begin{gathered} \boxed{ \sf \pink{ \tan \: a = \frac{p}{b} = \frac{5k}{12k} = {\frac{5}{12} }}} \\ \end{gathered}

tana=

b

p

=

12k

5k

=

12

5

\boxed{ \sf \red{ \cos \: a = \frac{b}{h} = \frac{12k}{13k} = \frac{12}{13}} }

cosa=

h

b

=

13k

12k

=

13

12

}pleasemarkasbrainliest............