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

(i) cos ∠DBC

(ii) cot ∠DBA

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$

⇒ AC = 13

Let ∠CBD be x°.

So, ∠DBA = 90° - x°.

In Δ DAB, according to angle sum property

⇒ ∠ DAB + ∠ADB + ∠DBA = 180°

⇒ ∠ DAB + 90° + (90° - x°) = 180°

⇒ ∠ DAB + 180° - x° = 180°

⇒ ∠ DAB - x° = 0

⇒ ∠ DAB = x°

From figure,

∠ DAB = ∠ CAB = x°

∴ ∠ CBD = ∠ CAB = x°

(i) cos ∠DBC = cos ∠CAB = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{12}{13}$

Hence, cos ∠DBC = $\dfrac{12}{13}$.

(ii) In Δ BCD, according to angle sum property

⇒ ∠ DBC + ∠DCB + ∠CDB = 180°

⇒ ∠ DCB + x° + 90° = 180°

⇒ ∠ DCB = 180° - 90° - x°

⇒ ∠ DCB = 90° - x°

From figure,

∠ DCB = ∠ ACB = 90° - x°

∴ ∠ DBA = ∠ ACB = 90° - x°

cot ∠DBA = cot ∠ACB = $\dfrac{Base}{Perpendicular}$

$= \dfrac{BC}{AB}\\[1em] = \dfrac{5}{12}$

Hence, cot ∠DBA = $\dfrac{5}{12}$.