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

Answer:

see explanation

Step-by-step 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 = $\sqrt{625}$ = 25

Then

sinA = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{7}{25}$

cosA = $\frac{adjacent}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{24}{25}$

sinC = $\frac{AB}{AC}$ = $\frac{24}{25}$

cosC = $\frac{BC}{AC}$ = $\frac{7}{25}$