In `DeltaABC,` `angleC=90^(@) and coses A=(13)/(12)`, find the values of all other trigonometric ratios for `angleA`,

Explanation

We know that ,
`"cosc"A=("hypotenuse")/("perpendicular")=(13)/(12)`
Now , construct a `DeltaABC` , from Pythagoras theorem
`AC^(2)+BC^(2)=AB^(2)`
`rArr" " AC^(2)=AB^(2)-BC^(2)=(13k)^(2)-(12k)^(2)`
`" " =169k^(2)-144k^(2)=25k^(2)`
`rArr" " AC=5k`
Now , `sinA=("perpendicular")/("hypotenuse")=(BC)/(AB)=(12k)/(13k)=(12)/(13)`
`cosA=("base")/("hypotenuse")=(AC)/(AB)=(5k)/(13k)=(5)/(13)`
`tanA=("perpendicular")/("base")=(BC)/(AC)=(12k)/(5k)=(12)/(5)`
`secA=("hypotenuse")/("base")=(BA)/(AC)=(13k)/(5k)=(13)/(5)`
`cotA=("base")/("perpendicular")=(AC)/(BC)=(5k)/(12k)=(5)/(12)`