Question

In the diagram, given below,triangle ABC is right-angled at B and BD is perpendicular to AC.Find:
i) cos angle DBC
ii) cot angle DBA

Answer 1

Answer:

Cos∠DBC = 12/13

Cot ∠DBA  = 5/12

Step-by-step explanation:

In the diagram, given below,triangle ABC is right-angled at B and BD is perpendicular to AC.Find:

i) cos angle DBC

ii) cot angle DBA

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

Answer 2

Answer:

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