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

Answer 1

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.