Nice College Test Geometry Problem | Find the side lengths of the triangle | 2 Ways to Solve
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- Опубліковано 14 жов 2024
- Nice College Test Geometry Problem | Find the side lengths of the triangle | 2 Ways to Solve
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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).
Since the ratio of the sides of a triangle with angles 30,60and 90 degrees are 1:root3 :2, c will be twice a.
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
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
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)
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.
In the original problem, ∠A = θ, ∠B = 2θ, & ∠C = 2θ.
Now solve that for the EXACT answers. Santa Maria!
.
.
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
.
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
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.
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
?
6/8/10 do not satisfy 30/60/90angle conditions
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. 🙏
This would be a lot easier if these college kids had just memorized the 30-60-90 triangle when they were in high school.
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)
3^(1/4)=4th root of 3
and 4* 3^3/4 = 9. 11803
what s the second solution?
The last few seconds, finding b and c using tan and sin.
Sir 🙏 after finding out a,b and c why again tan30 , Sin 30..
That is the 2nd way to find b and c after finding a.
🙏
Took me 10 seconds, 30-60-90 triangle.
Nice
θ=30,ovviamente...poi sappiamo che A=1/2absinα...i lati risultano √√3√48,√48/√√3,2√48/√√3
C = 2 times a
Задача не очень сложная решается в уме
A simple 6-8-10 right triangle. What is this fuss about?