In a triangle ABC , let D be the m.p of BC and angleADB 45degree and ACD 30 degree ,determine angle BAD?

Answer: <BAD = 30°

Solution:

i) In triangle ADB, by angle sum property, <DBA + <BAD = 180° - 45° = 135°

==> <DBA = 135° - <BAD

[Let this be y = 135° - x for convenience; and we need to find x = ?]

ii) In triangle CAD, <C = 30° (Given); <CDA = 135° (By liner pair propery)

==> So <CAD = 15°

iii) By sine law of triangles,

From triangle BAD, BD/sin(x) = AB/sin(45°) = AD/sin(y)

From triangle CAD, CD/sin(15°) = AC/sin(135°) = AD/sin(30°)

So dividing one by other, we get, sin(x)/sin(15°) = sin(y)/sin(30°)

==> sin(y)/sin(x) = sin(30°)/sin(15°) = 2sin(15°)cos(15°)/sin(15°) = 2*cos(15°)

Substituting for y from step (i):

sin(135 - x)/sin(x) = 2*cos(15)

{sin(135)*cos(x) - cos(135)*sin(x)}/sin(x) = 2*cos(15)

{sin(45)*cos(x) + sin(45)*sin(x)}/sin(x) = 2*cos(15) [applying supplementary and complementary angle relations]

==> {cos(x) + sin(x)}/sin(x) = 2*cos(15)/sin(45) = 2*{(√3 + 1)/2√2}/(1/√2) = √3 + 1

==> cot(x) + 1 = √3 + 1

==> cot(x) = √3

So x, that is <BAD = 30 deg.

if you have more maths question use mathqu homework support app in app store this is the best app to get answers from live tutors who write step by step maths answers search for mathqu in app store

Answer: <BAD = 30°

Solution:

i) In triangle ADB, by angle sum property, <DBA + <BAD = 180° - 45° = 135°

==> <DBA = 135° - <BAD

[Let this be y = 135° - x for convenience; and we need to find x = ?]

ii) In triangle CAD, <C = 30° (Given); <CDA = 135° (By liner pair propery)

==> So <CAD = 15°

iii) By sine law of triangles,

From triangle BAD, BD/sin(x) = AB/sin(45°) = AD/sin(y)

From triangle CAD, CD/sin(15°) = AC/sin(135°) = AD/sin(30°)

So dividing one by other, we get, sin(x)/sin(15°) = sin(y)/sin(30°)

==> sin(y)/sin(x) = sin(30°)/sin(15°) = 2sin(15°)cos(15°)/sin(15°) = 2*cos(15°)

Substituting for y from step (i):

sin(135 - x)/sin(x) = 2*cos(15)

{sin(135)*cos(x) - cos(135)*sin(x)}/sin(x) = 2*cos(15)

{sin(45)*cos(x) + sin(45)*sin(x)}/sin(x) = 2*cos(15) [applying supplementary and complementary angle relations]

==> {cos(x) + sin(x)}/sin(x) = 2*cos(15)/sin(45) = 2*{(√3 + 1)/2√2}/(1/√2) = √3 + 1

==> cot(x) + 1 = √3 + 1

==> cot(x) = √3

So x, that is <BAD = 30 deg.

Note: This is my own original solution. Kindly be aware of one other member who regularly copies other solutions and present the same as of his/her own.