Question

In a right angle triangle ABC angle B=90⁰ and AB+BC= 31cm AB-BC=17 cm the find sinA,cosA,sinC,cosC.​

Answer 1

see explanation

Explanation:

Given

AB + BC = 31

AB - BC = 17

Add the 2 equations

2AB = 48 ( divide both sides by 2 )

AB = 24

Substitute AB = 24 into the first equation

24 + BC = 31 ( subtract 24 from both sides )

BC = 7

Using Pythagoras' identity to find the hypotenuse AC

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 ( take square root of both sides )

AC =
`√(625)` = 25

Then

sinA =
`(opposite)/(hypotenuse)` =
`(BC)/(AC)` =
`(7)/(25)`

cosA =
`(adjacent)/(hypotenuse)` =
`(AB)/(AC)` =
`(24)/(25)`

sinC =
`(AB)/(AC)` =
`(24)/(25)`

cosC =
`(BC)/(AC)` =
`(7)/(25)`