ΔABC is a right angled at B and BD is perpendicular to AC. Find cos angle CBD. And also find cot angle ABD.
Answers
(a) ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R
It is given that
AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x⁰ and ∠SAB = y⁰
By Geometry ∆ARS and ∆ABC are similar
AR/AB = RS/BC
Substituting the values
AR/18 = 5/7.5
By further calculation
AR = (5 × 18)/7.5 = (1 × 18)/1.5
Multiply both numerator and denominator by 10
AR = (18 × 10)/15
AR = (10 × 6)/5
AR = (2 × 6)/1 = 12
So we get
RB = AB – AR
RB = 18 – 12 = 6
In right angled ∆ABC
Using Pythagoras theorem
AC² = AB² + BC²
Substituting the values
AC² = 182 + 7.52
By further calculation
AC² = 324 + 56.25 = 380.25
AC = √380.25 = 19.5 cm
(i) In right angled ∆BSR
tan x⁰ = perpendicular/base
tan x⁰ = RB/RS = 6/5
(ii) In right angled ∆ASR
sin y⁰ = perpendicular/hypotenuse
Using Pythagoras theorem
AS2 = 122 + 52
By further calculation
AS2 = 144 + 25 = 169
AS = √169 = 13 cm
So we get
sin y⁰ = RS/AS = 5/13
(b) We know that
∆ABC is right angled at B and BD is perpendicular to AC
In right angled ∆ABC
Using Pythagoras theorem
AC² = AB² + BC²
Substituting the values
AC² = 12² + 5²
By further calculation
AC2 = 144 + 25 = 169
So we get
AC² = (13)²
AC = 13
By Geometry ∠CBD = ∠A and ∠ABD = ∠C
(i) cos ∠CBD = cos ∠A = base/hypotenuse
In right angled ∆ABC
cos ∠CBD = cos ∠A = AB/AC = 12/13
(ii) cos ∠ABD = cos ∠C = base/perpendicular
In right angled ∆ABC
cos ∠ABD = cos ∠C = BC/AB = 5/12
(a) ∆ABC is right-angled at B, ∆BSC is right-angled at S and ∆BRS is right-angled at R
It is given that
AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x⁰ and ∠SAB = y⁰
By Geometry ∆ARS and ∆ABC are similar
AR/AB = RS/BC
Substituting the values
AR/18 = 5/7.5
By further calculation
AR = (5 × 18)/7.5 = (1 × 18)/1.5
Multiply both numerator and denominator by 10
AR = (18 × 10)/15
AR = (10 × 6)/5
AR = (2 × 6)/1 = 12
So we get
RB = AB – AR
RB = 18 – 12 = 6
In right angled ∆ABC
Using Pythagoras theorem
AC² = AB² + BC²
Substituting the values
AC² = 182 + 7.52
By further calculation
AC² = 324 + 56.25 = 380.25
AC = √380.25 = 19.5 cm
(i) In right angled ∆BSR
tan x⁰ = perpendicular/base
tan x⁰ = RB/RS = 6/5
(ii) In right angled ∆ASR
sin y⁰ = perpendicular/hypotenuse
Using Pythagoras theorem
AS2 = 122 + 52
By further calculation
AS2 = 144 + 25 = 169
AS = √169 = 13 cm
So we get
sin y⁰ = RS/AS = 5/13
(b) We know that
∆ABC is right angled at B and BD is perpendicular to AC
In right angled ∆ABC
Using Pythagoras theorem
AC² = AB² + BC²
Substituting the values
AC² = 12² + 5²
By further calculation
AC2 = 144 + 25 = 169
So we get
AC² = (13)²
AC = 13
By Geometry ∠CBD = ∠A and ∠ABD = ∠C
(i) cos ∠CBD = cos ∠A = base/hypotenuse
In right angled ∆ABC
cos ∠CBD = cos ∠A = AB/AC = 12/13
(ii) cos ∠ABD = cos ∠C = base/perpendicular
In right angled ∆ABC
cos ∠ABD = cos ∠C = BC/AB = 5/12
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