For the triangle Below BD= 6

Question

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)

Answer 1

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.