Or, sine of angle ABD (75 degrees) =.9659 . AD/.9659 = AB. Thus, AB = 3.1059 . Cosine of AB (BD) at 75 degrees is .2588 of AB or .8036. BD is thus = .8036. BD minus BC, gives DC of 5.1961. Tangent of angle ACD is AD/DC or .5773 which is 30 degrees. Short and sweet.
The second method shows how useful-in an indirect way an isosceles obtuse triangle with two 15 degree angles really is. And that is probably the clever geometric equivalent is the tangent identity that was applied in the first method.
To a point E between DC, draw AE such that ∠BAE =75°, implying ∆AEB is Isosceles with ∠AEB =30°. Since, AE*Sin30= AD=3, AE=6, which is also equal to BE. Hence ∆AEB is Congruent to ∆ACB and θ=∠AEB=30° If E was to the right of DC, exactly same argument will hold.
Use Math Booster shape in 07:00 Let BD=x. In right triangle BE=2x. ΔABE is isosceles => AE=BE=2x DE=AD-AE=> DE=3-2x Apply Pythagoras theorem in right triangle BDE => BD²+DE²=BE² => x²+(3-2x)²=(2x)² => x²-12x+9=0 => x=6-3√3 or ( x=6+3√3 is rejected)** Now DC=BC-BD=6-x=6-(6-3√3 ) => DC=3√3 =>DC²=27 Apply Pythagoras theorem in right triangle ADC => AC²=AD²+DC² => AC²=3²+27 => AC=6. AC=BC=6 => ΔABC is isosceles ……… θ = 30° ** DE=3-2x >0 => x
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Or, sine of angle ABD (75 degrees) =.9659 . AD/.9659 = AB. Thus, AB = 3.1059 . Cosine of AB (BD) at 75 degrees is .2588 of AB or .8036. BD is thus = .8036. BD minus BC, gives DC of 5.1961. Tangent of angle ACD is AD/DC or .5773 which is 30 degrees. Short and sweet.
Excelente!
BD = 3cot(75)
DC = 6 - 3cot(75) = 3sqrt(3)
tan(Theta) = 3/(3sqrt(3)) = sqrt(3)/3
Theta = 30
The second method shows how useful-in an indirect way an isosceles obtuse triangle with two 15 degree angles really is. And that is probably the clever geometric equivalent is the tangent identity that was applied in the first method.
To a point E between DC, draw AE such that ∠BAE =75°, implying ∆AEB is Isosceles with ∠AEB =30°. Since, AE*Sin30= AD=3,
AE=6, which is also equal to BE.
Hence ∆AEB is Congruent to ∆ACB
and θ=∠AEB=30°
If E was to the right of DC, exactly same argument will hold.
Parabéns pelo trabalho!!! 👏🏻
Use Math Booster shape in 07:00
Let BD=x. In right triangle BE=2x.
ΔABE is isosceles => AE=BE=2x
DE=AD-AE=> DE=3-2x
Apply Pythagoras theorem in right triangle BDE => BD²+DE²=BE² =>
x²+(3-2x)²=(2x)² => x²-12x+9=0 => x=6-3√3 or ( x=6+3√3 is rejected)**
Now DC=BC-BD=6-x=6-(6-3√3 ) => DC=3√3 =>DC²=27
Apply Pythagoras theorem in right triangle ADC => AC²=AD²+DC² =>
AC²=3²+27 => AC=6.
AC=BC=6 => ΔABC is isosceles ……… θ = 30°
** DE=3-2x >0 => x
I solved it by the 1st method. It's more direct.
(3)^2H/A/ASino°+ (6)^2H/A/BCoso°={9H/A/ASino°+36H/A/Bcoso°}= 45H/A/ASino°BCoso° {75°A+75°B}= 150°AB {150°AB+30°C}=180°ABC 180°ABC/45H/A/ASino°BCoso°=2:.20 H/A/ASino°BCoso° 2^1.5^4 2^1.5^1^4 2^1.1^12^2 1^1.1^2 1^2 (H/A/ASino°BCoso°ABC ➖ 2H/A/ASino°BCoso°ABC+1).
4:45
You missaid that 6 - x = 2√3 when 6 - x = 3√3.
AB*cos75 = x AC*cos teta = 6-x AC*sin teta = 3 > tg teta = 3/(6-x) ... teta = 30
φ = 30° → sin(3φ) = 1; ∆ ABC → AB = a; BC = BD + CD = c + (6 -c); AC = b; sin(BDA) = 1
AD = 3; BCA = θ; BDA = 5φ/2 → DAB = φ/2 →
sin(φ) = 1/2 → cos(φ) = √3/2 → sin(φ/2) = √((1/2)(1 - cos(φ))) = (√2/4)(√3 - 1)
cos(φ/2) = √((1/2)(√3 + 1))) = (√2/4)(√3 + 1) →
tan(φ/2) = sin(φ/2)/cos(φ/2) = 2 - √3 = c/3 → c = 3(2 - √3) →
6 - c = 3√3 → b = 6 → sin(θ) = 3/b = 1/2 → θ = φ
tan(75°) = DA/BD
tan(45°+30°) = 3/BD
(tan(45°)+tan(30°))/(1-tan(45°)tan(30°)) = 3/BD
(1+1/√3)/(1-1/√3) = 3/BD
(1+1/√3)BD = 3 - √3
BD = (3-√3)/(1+1/√3)
BD = (3-√3)(1-1/√3)/(1+1/√3)(1-1/√3)
BD = (3-√3-√3+1)/(1-1/3)
BD = (4-2√3)(3/2) = 6 - 3√3
tan(θ) = DA/CD
tan{θ) = 3/(6-(6-3√3)) = 3/3√3 = 1/√3
θ = tan⁻¹(1/√3) = 30°