Question

if cosA=5/13 evaluate: 13sinA+5secA/5sinA+6cosecA​

Answer 1

since,

sinA=p/h

cosA=b/h

tanA=p/b

therefore now,

given that, cosA=5/13 which is equal to b/h

there for b=5,h=13

now we find 'p'

h²=p²+b²

13²=p²+5²

p²=13²-5²

p²=169-25

p²=144

p=√144

p=12

now, putting the values of p,h,b

1. 13sinA+5secA/5sinA+6cosecA
2. 13×12/13+5×13/5

/

5×12/13+6×13/12

25

/

120+169

/

26

25×26/289

650/289