Question

Hello there ^ _ ^

I am not sure how to solve this thing - May you help me please ? Thanks~

Answer 1

the answer is : ( D )

Answer 2

`cos(30^o)= (adjacent)/(hypotenuse) = (x)/(16 √(3) ) \ \ \to \\x=16 √(3)*cos(30^o) =16 √(3)* \cfrac{ √(3) }{2} =8*3=24 \\ \\ \\ sin(30^o)= (opposite )/(hypotenuse) =(y)/(16 √(3) ) \ \ \to \\y=16 √(3)*sin(30^o) =16 √(3)* \cfrac{1 }{2} =8 √(3)`

Answer: C.