Question

Please help! urgently

Answer 1

Answer:

sinC = $\frac{4\sqrt{41} }{41}$

Step-by-step explanation:

We require to find the side DE before finding sinC

Using Pythagoras' identity in the right triangle

DE² + CE² = CD²

DE² + 5² = ($\sqrt{41}$ )²

DE² + 25 = 41 ( subtract 25 from both sides )

DE² = 16 ( take the square root of both sides )

DE = $\sqrt{16}$ = 4 , then

sinC = $\frac{opposite}{hypotenuse}$ = $\frac{DE}{CD}$ = $\frac{4}{\sqrt{41} }$ , rationalise the denominator

sinC = $\frac{4}{\sqrt{41} }$ × $\frac{\sqrt{41} }{\sqrt{41} }$ = $\frac{4\sqrt{41} }{41}$