Answer:
[tex]\sin{B}=0.6[/tex]
Step-by-step explanation:
We remember that the sine of any angle in a right triangle is the side opposite that angle divided by the hypotenuse. So, [tex]\sin{B}=\frac{CD}{BD}=\frac{CD}{\sqrt{69}}[/tex].
But we don't know CD!
Aha! We can use the pythagorean theorem! Because [tex]\triangle BCD[/tex] is right, we have that [tex]BC^2+CD^2=BD^2[/tex], so [tex]44+CD^2=69[/tex].
Solving gives [tex]CD=\pm 5[/tex]. However, because CD is a side length, it must be positive. So, we have [tex]CD=5[/tex].
Plugging into our formula for [tex]\sin{B}[/tex] gives [tex]\sin{B}=\frac{5}{\sqrt{69}}=0.601929265...\approx \boxed{0.6}[/tex] rounded to the nearest hundredth.
Note: Find [tex]\cos{D}[/tex]. Note that(if you've done it correctly) it equals [tex]\sin{B}[/tex]. Coincidence? If not, try and prove it!
Hint for note: It's not a coincidence. One of the first facts you'll learn is that [tex]\sin{\theta}=\cos{90-\theta}[/tex]. Try and prove this for acute angles in a triangle!
