If SEC A=13/12 Calculate Sin A and Cot A. (A is an acute angle.) | Bzziii.com

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If sec A=`13/12`, calculate sin A and cot A. (A is an acute angle.)









 Finding cos θ 

 cos θ = `"1"/"sec θ "`

 cos θ = `\frac{1}{\frac{13}{12}}`

 cos θ = `\frac{12} {13}`


`"Side adjacent to ∠θ "/"Hypotenuse"` = `\frac{12} {13}`

`"AB"/"BC"` = `12/13`


Let, AB = 12x

& AC = 13x


Using Pythagoras theorem to find BC

If sec A=13/12 calculate sin A and cot A. (A is an acute angle.)

`"(Hypotenuse)"^2` = `"(Height)"^2` +  `"(Base)"^2`

`"AC"^2` = `"AB"^2` + `"BC"^2` 

`"(13x)"^2` = `"(12x)"^2` +  `"(BC)"^2`

 `"(BC)"^2` = `"(13x)"^2` - `"(12x)"^2` 

 `"(BC)"^2` = 169`"x"^2` - 144`"x"^2`

 `"(BC)"^2` = 25`"x"^2`

BC = `\sqrt{25"x"^{2}}`

BC = `\sqrt{5^2x^{2}}`

BC = 5x



Now, sin θ = `"BC"/"AC"`

= `"5x"/"13x"`

= `"5"/"13"`


cos θ = `"12"/"13"`


tan θ = `"BC"/"AB"`

= `"5x"/"12x"`

= `"5"/"12"`


cot θ = `"1"/"tan θ "`

= `\frac{1}{\frac{5}{12}}`

= `12/5`

Thus, sin θ = `"5"/"13"`  and  cot θ = `"12"/"5"`






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