For the triangle Below BD= 6
CD= 20 and BC= 2 √109
Solve for sin(B) cos(B) and tan(B) rounded to nearest tenth (note sides not drawn to scale)
To solve for sin(B), cos(B), and tan(B) in the triangle above, we first need to find the length of side BC using the Pythagorean theorem.
BC^2 = BD^2 + CD^2
BC^2 = 6^2 + 20^2
BC^2 = 36 + 400
BC^2 = 436
BC = √436
BC = 2√109 (Given)
Now, we can use the trigonometric ratios to solve for sin(B), cos(B), and tan(B):
sin(B) = opposite / hypotenuse = BC / CD = (2√109) / 20 ≈ 0.447
cos(B) = adjacent / hypotenuse = BD / CD = 6 / 20 = 0.3
tan(B) = opposite / adjacent = BC / BD = (2√109) / 6 ≈ 1.2
Therefore, sin(B) ≈ 0.4, cos(B) = 0.3, and tan(B) ≈ 1.2 rounded to the nearest tenth.