Question

In this triangle, the product of sin B and tan C is
and the product of sin Cand tan B is

Answer 1

`(c)/(a)` and
`(b)/(a)`

Explanation:

sinB =
`(opposite)/(hypotenuse)` =
`(AC)/(BC)` =
`(b)/(a)`

tanC =
`(opposite)/(adjacent)` =
`(AB)/(AC)` =
`(c)/(b)`

Thus

sinB tanC =
`(b)/(a)` ×
`(c)/(b)` ( cancel b on numerator/ denominator )

=
`(c)/(a)`

---------------------------------------------------------------------------

sinC =
`(opposite)/(hypotenuse)` =
`(AB)/(BC)` =
`(c)/(a)`

tanB =
`(opposite)/(adjacent)` =
`(AC)/(AB)` =
`(b)/(c)`

Thus

sinC tanB =
`(c)/(a)` ×
`(b)/(c)` ( cancel c on numerator/ denominator )

=
`(b)/(a)`