Question

Find the sine of ZD.
F
2√15
D
2-√15
E

Answer 1

`{ \underline{ \underline{ \mathfrak{ \: answer : \: \: { \tt{ \sin(D) = ( √(2) )/(2) \: \: }} }}}}`

Explanation:

`{ \rm{ \sin( \theta) = (opposite)/(hypotenuse) }} \\`

• Theta is D
• opposite is 2sqrt(15)
• hypotenuse is h

`{ \tt{h = \sqrt{ {(opposite)}^(2) + {(adjacent)}^(2) } }} \\ { \tt{h = \sqrt{ 2{(2 √(15)) }^(2) } }} \\ { \tt{h = √(2(4 * 15)) }} \\ { \tt{h = √(120) }}`

Therefore:

`{ \tt{ \sin(D) = (2 √(15) )/( √(120) ) }} \\ \\ { \tt{ \sin(D) = (2 √(15) )/(2 √(15 * 2) ) }} \\ \\ { \tt{ \sin(D) = (1)/( √(2) ) }} \\ \\ { \boxed{ \tt{ \sin(D) = ( √(2) )/(2) }}}`