Question

given : `15 cotA = 8 ` find `sinA and secA`

Answer 1

`15cotA=8rArrcotA=(8)/(15)` Construct `DeltaABC` , right -angled at B such that
`AB=8kandBC=15k`
In `DeltaABC` ,
From Pythagoras theorem
`AC^(2)=AB^(2)+BC^(2)=(8k)^(2)+(15k)^(2)`
`=64k^(2)+225k^(2)=289k^(2)`
`rArrAC=17k`
Now `sinA=(BC)/(AB)=(15k)/(17k)=(15)/(17)`
`secA=(AC)/(AB)=(17k)/(8k)=(17)/(8)`