Question

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

1. cos ∠DBC
2. cot ∠DBA

Answer 1

Consider the given figure :

Since the triangle is a right-angled triangle, so using Pythagorean Theorem

AC² = 5² + 12²

AC² = 25 + 144 + 169

AC = 13

In ΔCBD and ΔCBA, the ∠C is common to both the triangles, ∠CDB = ∠CBA = 90° so therefore ∠CBD = ∠CAB.

Therefore, ΔCBD and ΔCBA are similar triangles according to AAA Rule 
So
`"AC"/"BC" = "AB"/"BD"`

`(13)/(5) = ( 12)/"BD"`

`"BD = (60)/(13)`

In ΔDBA,

AB = 12; BD = `60/13`

AD = 13 - x

= `13 - 25/13 = (169 - 25)/13 = 144/13`

(i) cos ∠DBC = `"base"/"hypotenuse" = "BD"/"BC" = (60/13)/(5) =(12)/(13)`

(ii) cot ∠DBA =`"base"/"perpendicular" = "BD"/"AB" = (60/13)/(144/13) =(5)/(12)`

Answer 2

Since the triangle ABC is a right-angled triangle, so using Pythagorean Theorem,

AC² = BC² + AB²

AC² = 5² + 12²

AC² = 25 + 144

AC² = 169

AC² = `sqrt169`

AC = 13

In ΔDBC, using Pythagorean Theorem,

BC² = CD² + BD²

(5)² = x² + BD²

25 - x² = BD²

BD² = 25 - x²     ...(i)

In ΔDBA, using Pythagorean Theorem,

BA² = DA² + BD²

(12)² = (13 - x)² + BD²

144 = (13 - x)² + BD²

144 = 169 + x² - 26x + BD²

144 - 169 - x² + 26x = BD²

BD² = - 25 - x² + 26x   ...(ii)

From equation (i) and (ii)

25 - x² = - 25 - x² + 26x

25 + 25 = 26x

x = `50/26`

x = `25/13`

In ΔBDC,

BC = 5 ; CD = `25/13`

BD² = 25 - x²

= 25 - `(25/13)^2`

= 25 - `625/169`

= 25 `(1 - 25/169)`

= `25 ((169 - 25)/169)`

BD = `sqrt(25 (144/169))`

BD = `(5 xx 12)/13`

BD = `60/13`

In ΔDBA,

AB = 12; BD = `60/13`

AD = 13 - x

= `13 - 25/13 = (169 - 25)/13 = 144/13`

(i) cos ∠DBC = `"Base"/"hypotenuse" = (60/13)/5 = 60/65 = 12/13`

(ii) cot ∠DBA `= "Base"/"perpendicular" = (60/13)/(144/13) = 60/144 = 10/24 " i.e.," 5/12`