Question

Construct a triangle ABC in which AB = 6 cm, _A = 75° and C = 45°
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Answer 1

In ∆ ABC A+B+C=180° , putting A=75° and. C=60°

75°+B+60°=180° => B= 45°. , let angle. ABD=x° then angle DBC= (45-x).

Given that :- area of ∆ BAD = √3.area of∆ BCD.

1/2.AB.BD.sin x = √3.1/2.BC.BD.sin(45-x)

or. AB.sin x. = √3.BC.sin(45-x)………………(1)

In ∆ABC by Sine rule:-

AB/sin60 = BC/sin 75…………………. (2)

Dividing eqn. (1). by (2)

sin60.sin x=√3sin75.sin(45-x)

or. √3/2.sin x= √3.sin75.sin(45-x).

or. sin x=2.sin75.sin(45-x)

or. sin x=cos(75–45+x)-cos(75+45-x)

or. sin x =cos(30+x) -cos (120-x)

or. sin x=cos(30+x) -cos{90+(30-x)}

or. sin x= cos(30+x) +sin(30-x)

or. sin x=cos30.cos x-sin30.sin x+sin30.cos x-cos30.sin x

or. sin x=√3/2.cos x-1/2.sin x+1/2cos x-√3/2.sin x

or. 2.sin x=√3.cos x-sin x+cos x-√3.sin x

or. (2+1+√3).sin x=(√3+1).cos x

or. sin x/cos x= (√3+1)/(3+√3) = (√3+1)/{√3.(√3+1)} = 1/√3

or. tan x= 1/√3. or. tan x=tan30°. => x=30°. answer

I hope it helps you mate ✨