In the figure (2) given below, ∆ABC is right-angled at B and BD is perpendicular to AC. Find :

(i) cos ∠CBD

(ii) cot ∠ABD

In right angle ∆ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13.

Let ∠CBD = x.

∠DBA = 90° - x

In ∆DAB,

⇒ ∠DAB + ∠ADB + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠DAB + 90° + 90° - x = 180°

⇒ ∠DAB = 180° - 180° + x

⇒ ∠DAB = x.

From figure,

∠DAB = ∠CAB = x

∴ ∠CBD = ∠CAB = x

(i) cos ∠CBD = cos ∠CAB

= $\dfrac{AB}{AC} = \dfrac{12}{13}$.

Hence, cos ∠CBD = $\dfrac{12}{13}$.

(ii) In ∆BCD,

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [Angle sum property of triangle]

⇒ ∠DCB + x + 90° = 180°

⇒ ∠DCB = 180° - 90° - x

⇒ ∠DCB = 90° - x.

From figure,

∠DCB = ∠ACB =90° - x

∴ ∠ABD = ∠ACB = 90° - x

∴ cot ∠ABD = cot ∠ACB

= $\dfrac{BC}{AB} = \dfrac{5}{12}$.

Hence, cot ∠ABD = $\dfrac{5}{12}$.