Question

Find the sine of ZD.
F
2√15
D
2-√15
E

Answer 1

Answer:

${ \underline{ \underline{ \mathfrak{ \: answer : \: \: { \tt{ \sin(D) = \frac{ \sqrt{2} }{2} \: \: }} }}}}$

Step-by-step explanation:

${ \rm{ \sin( \theta) = \frac{opposite}{hypotenuse} }}$

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

${ \tt{h = \sqrt{ {(opposite)}^{2} + {(adjacent)}^{2} } }} \\ { \tt{h = \sqrt{ 2{(2 \sqrt{15}) }^{2} } }} \\ { \tt{h = \sqrt{2(4 \times 15)} }} \\ { \tt{h = \sqrt{120} }}$

Therefore:

${ \tt{ \sin(D) = \frac{2 \sqrt{15} }{ \sqrt{120} } }} \\ \\ { \tt{ \sin(D) = \frac{2 \sqrt{15} }{2 \sqrt{15 \times 2} } }} \\ \\ { \tt{ \sin(D) = \frac{1}{ \sqrt{2} } }} \\ \\ { \boxed{ \tt{ \sin(D) = \frac{ \sqrt{2} }{2} }}}$