For the triangle Below BD 6 CD=20 and BC=2v/109 solve for sin(B) cos(B) and tan(B) rounded to nearest tenth (note sides not drawn to scale
To find sin(B), cos(B) and tan(B) in the triangle given, we can use the Law of Cosines:
cos(B) = (a^2 + c^2 - b^2) / 2ac
Where a is the side opposite angle A (BC), b is the side opposite angle B (BD), and c is the side opposite angle C (CD).
Given that BD = 6, CD = 20, and BC = 2√109, we can plug in these values to find cos(B):
cos(B) = ([2√109]^2 + 6^2 - 20^2) / (2 * 2√109 * 6)
cos(B) = (436 + 36 - 400) / (12√109)
cos(B) = 72 / 12√109
cos(B) ≈ 0.620
Now, we can find sin(B) using the Pythagorean theorem:
sin(B) = BD / BC
sin(B) = 6 / 2√109
sin(B) = 3 / √109
sin(B) ≈ 0.287
Finally, we can find tan(B) using the relationship between sine and cosine:
tan(B) = sin(B) / cos(B)
tan(B) = 0.287 / 0.620
tan(B) ≈ 0.463
Therefore, sin(B) ≈ 0.3, cos(B) ≈ 0.6, and tan(B) ≈ 0.5.