Question

triangle ABC is right angled triangle at B and BD is perpendicular to AC . BA is 12 and BC is 5 . find cos angle DBC and cot angle DBA​

Answer 1

Answer:

Cos∠DBC = 12/13

Cot ∠DBA = 5/12

Step-by-step explanation:

ΔBDC & ΔABC

∠B = ∠D = 90°

∠C = ∠C Common angle

=> ∠DBC = ∠A

ΔBCA ≅ ΔDCB

Cos∠A = AB/AC

AC² = AB² + BC² = 12² + 5² = 169

=> AC = 13

Cos∠A = 12/13

=> Cos∠DBC = 12/13

∠DBA = 90 - ∠DBC

=> ∠DBA = 90 - A

Cot ∠DBA = cot (90 - A) = tan ∠A

tan ∠A = P/B = 5/12

Cot ∠DBA = 5/12