Nice College Test Geometry Problem | Find the side lengths of the triangle | 2 Ways to Solve

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Nice College Test Geometry Problem | Find the side lengths of the triangle | 2 Ways to Solve
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Комментарии : 29

@anandharamang3289 5 месяцев назад

This method will be so easy. Angle theta =30°, hence it is right angle triangle of 30, 60, 90 and side ratio of 1,√3, 2 respectively. Area is 1/2*1k*√3k=(√3/2)k^2= 24 So, k= √48/√3= 4*3^(1/4) And a=k*1 , b=k*√3 , c = k*2 a= 4*3^(1/4) ; b= 4*3^(3/4); and c= 8*3^(1/4).

@johnbrennan3372 5 месяцев назад

Since the ratio of the sides of a triangle with angles 30,60and 90 degrees are 1:root3 :2, c will be twice a.

@devondevon4366 5 месяцев назад

30, 60, 90 right triangle ab = 48 a *asqrt 3 = 48 a^2 = 48 sqrt 3/3 a^2 = 16 sqrt 3 a = 5.2643 c = 5.2643 x 2= 10.5286 d (Pythagorean) = 9.11803

@Ramkabharosa 5 месяцев назад

In the original problem, ∠A = θ, ∠B = 2θ, & ∠C = 2θ. Now solve that for the EXACT answers. Santa Maria! .

@PrithwirajSen-nj6qq Месяц назад

Second Method 1/2absin90=1/2bcsin30=1/2casin60=24 Here we get ab, bc, ca Then (abc)^2 Then (abc) 2/(ab) 2=c^2 Then value of c In this way value of a and c will be known

@PeterCooperUK Месяц назад

I went a different way from the angles. Used sin 30 and sin 60 with an assumed hypotenuse of 1 to work out the ratio of the two opposite sides then boiled it down to the sides with that ratio that multiply up to 48 (then pythag for the hypoteneuse). A bit less algebraic but the end numeric result was the same.

@holyshit922 5 месяцев назад

1. Add measures of angles t+2t+3t = 180 6t = 180 t = 30 so we have right triangle From basic trigonometry a/c = 1/2 b/c = sqrt(3)/2 2Area = ab so ab = 48 a = 1/2c b = sqrt(3)/2c sqrt(3)/4c^2 = 48 c^2 = 192/sqrt(3) c^2 = 64*sqrt(3) c = 8*3^(1/4) b = 4*3^(3/4) a = 4*3^(1/4)

@PrithwirajSen-nj6qq Месяц назад

AB=2x,BC=x AC =√3x Area of ABC 1/2*√3x*x=24 Here we get the value of x Then the sides are x, 2x, √3x will be known. This is time saving method. Thanks. Plz offer remarks

@bpark10001 5 месяцев назад

You get way too elaborate in calculating AB. In 30/60/90 triangle, lengths are 1/2/sqrt(3). So AB is double the length calculated for BC.

@elizabethlasseter609 5 месяцев назад

This would be a lot easier if these college kids had just memorized the 30-60-90 triangle when they were in high school.

@allalwadi7063 5 месяцев назад

. Let angle theta = t t + 2t +3t =180° --> t=30° The triangle is 30° , 60° , 90° . ½ab =24 --> ab =48 , b= 48/a c = 2a ( since sin30° = ½ ) . a² +b²= c² a² + 48²/a² = 4a² 3a² - 48²/a² = 0 3a⁴ - 48² =0 a⁴ - 16(48) =0 (a² - 16✔️3) =0 a= 4 ⁴✔️3 , b= 12/ ⁴✔️3 , c= 8 ⁴✔️3 .

@Farhan-ph1sp 5 месяцев назад

Since interior angles are in A.P so, angles will be 30,60,90. If area is 24 it means base x height = 48 Then by hit and trial base and height will be 8 and 6 .. whereas hypotenuse will be 10 ..!! So the triangle is 30°-60°-90° and sides are 6-8-10

@cemildemir9317 5 месяцев назад

@mohanbhaipatel7266 4 месяца назад

6/8/10 do not satisfy 30/60/90angle conditions

@kaliprasadguru1921 4 месяца назад

Sir , when angle B is 60° then hypotenuse is twice the base or half of perpendicular . So why to go in for AB²= AC²+BC². 🙏

@bobbyheffley4955 5 месяцев назад

3^(1/4)=4th root of 3

@quigonkenny 2 месяца назад

Let AB = c, CA = b, and BC = a. θ + 2θ + 3θ = 180 6θ = 180 θ = 30° sin(θ) = a/c 1/2 = a/c c = 2a cos(θ) = b/c √3/2 = b/2a 2b = 2√3a b = √3a A = casin(2θ)/2 24 = (2a)a(√3/2)/2 48 = √3a² a² = 48/√3 = 16√3 a = √(16√3) = 4∜3 b = √3a = √3(4∜3) b = 4√(3√3) c = 2a = 2(4∜3) c = 8∜3 (a, b, c) = (4∜3, 4√(3√3), 8∜3)