Can you find the side lengths of the triangle? | (No Calculators!) |
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- Опубліковано 30 вер 2024
- Learn how to find the side lengths of the right triangle. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; area of a triangle formula. Step-by-step tutorial by PreMath.com
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Excellent working!😀🥂
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Can also be done using tan15= tan(45-30)=(sqroot 3-1)/(sqroot3+1)=|cb|\ |ab| etc. You then use the 3 eqs. You get |ab|=sqroot(4+2sqroot3) which works out as sqroot3 +1 etc.
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No calculator? No problem:
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..
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We know that sin(30°)=1/2 and that sin(2*x)=2*sin(x)*cos(x). With s=sin(15°) and c=cos(15°) we obtain:
1/2 = 2*s*c = 2*s*√(1 − s²)
1 = 4*s*√(1 − s²)
1 = 16*s²*(1 − s²) = 16*s² − 16*s⁴
16*s⁴ − 16*s² + 1 = 0
s⁴ − s² + 1/16 = 0
s² = (1/2) ± √(1/4 − 1/16) = (1/2) ± √(4/16 − 1/16) = (1/2) ± √3/4 = (2 ± √3)/4
First solution:
s² = (2 + √3)/4
s = √[(2 + √3)/4] = (√3 + 1)/(2√2)
c = √(1 − s²) = √[1 − (2 + √3)/4] = √[4/4 − (2 + √3)/4] = √[(2 − √3)/4] = (√3 − 1)/(2√2)
Second solution:
s² = (2 − √3)/2
s = √[(2 − √3)/4] = (√3 − 1)/(2√2)
c = √(1 − s²) = √[1 − (2 − √3)/4] = √[4/4 − (2 − √3)/4] = √[(2 + √3)/4] = (√3 + 1)/(2√2)
Since 0 < s=sin(15°) < c=cos(15°) we can conclude:
sin(15°) = (√3 − 1)/(2√2)
cos(15°) = (√3 + 1)/(2√2)
tan(15°) = (√3 − 1)/(√3 + 1) = (√3 − 1)²/[(√3 + 1)(√3 − 1)] = (3 − 2√3 + 1)/(3 − 1) = (4 − 2√3)/2 = 2 − √3
Now we are able to calculate the side lengths of the triangle:
BC*AB = 2*A(ABC) = 2cm²
BC/AB = tan(15°) = 2 − √3
BC² = (4 − 2√3)cm²
BC = (√3 − 1)cm
AB² = (2cm²)/(2 − √3) = (2cm²)(2 + √3)/[(2 − √3)(2 + √3)] = (2cm²)(2 + √3)/(4 − 3) = (4 + 2√3)cm²
AB = (√3 + 1)cm
AC² = AB² + BC² = (4 + 2√3)cm² + (4 − 2√3)cm² = 8cm²
AC = (2√2)cm
Best regards from Germany
Cool! Keep rocking!
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Very Nice...
Easy solution
Area = 1/2*x*y = 1
x*y = 2
AC*Sin(15)*AC*Cos(15) = 2
AC*AC Sin(15)*Cos(15) = 2
AC*AC*1/2*Sin(30) = 2
AC*AC = 8
AC =sqrt(8) ........1
AC*AC = x*x + y*y =( x+-y)*2 -+ 2xy = ( x+-y)*2 -+4 = 8
x+y = sqrt(8 + 4) = sqrt(12) positive x,y
x-y = sqrt(8 - 4) = sqrt(4) = -+2
2x = sqrt(12) -+2
x = sqrt(3) -+ 1
x is greater than y
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Respected sir,
you deserve 1 million subscribers 😊❤😊
Thank you so much 😀❤️
Alternative method using basic trig.
Let AC =h then x=hcos15 and y=hsin15
xy =(hcos15)(hsin15)=2
(h^2)(sin30)/2=2
h^2=8,so h=2sqrt2
x^2+y^2=8
(x+y)^2-2xy =8
(x+y)^2-4=8
x+y=2sqrt3 since x>y>0
Similarly x-y=2
x=sqrt3. +. 1
y = sqrt3. - 1
Use the trig identity:-
sin(A+B)=sinA*cosB+cosA*sinB...........................(1)
letA=45deg. and B=30deg.
sin45=1/sqrt2,cos45=1/sqrt2,sin30=1/2,cos30=sqrt3/2
substitute these values into (1),
sin(45+30)=(1/sqrt2)*(sqrt3/2)+(1/sqrt2)*(1/2)
sin(75)=(sqrt3+1)/(2*sqrt2).
sin75 = AB/AC,
We now have two values of your 15,75,90 triangle at11:39.
AB=(sqrt3 + 1) and AC= (2*sqrt2)
Use Pythagoras,
AC^2 = AB^2+BC^2
(2*sqrt2)^2=(sqrt3 + 1)^2+BC^2
8 =(3+1+2*sqrt3)+BC^2
BC^2=8-(4+2*sqrt3)=(4-2*sqrt3)=(sqrt3 -1)^2
BC^2= (sqrt3 -1)^2
BC= (sqrt3-1)
AB=(sqrt3+1)cm,BC=(sqrt3-1)cm and AC= (2*sqrt2) cm.
The ABC triangle is the base triangle to calculate the exact trig values of a 15,75.90 triangle.
If you are studying maths or physics get very familiar with all the base triangles and the trig.ratios
Thanks for the puzzle professor,this is a very important triangle.
Let AC = c
cos 15º = AB/AC
sin 15º =BC/AC
AB= C*cos15º and BC= C*sin15º
Area of the triangle 1/2*bh=1
1/2(AB*BC)=1
1/2 C cos 15º *Csin 15º=1
multiply by 4 on both sides
C^2 2sin15ºcos15º=4
C^2 sin(2*15º)=4
C^2 sin 30º=4
C^2=4/(1/2)=4*2=8
C= 2*sqrt2
from cos 30º
let a=15º
so cos 30º=cos 2a=(sqrt3)/2
by the double angle formula
cos 2a= 2*(cosa)^2-1
sqrt 3=2(2(cos a)^2 -1)
sqrt3=4 (cos a)^2 -2
(2+sqrt 3)/4=(cos a)^2
multiply the left side by 4 on top and bottom
(8+4sqrt3)/16
4sqrt3=sqrt48= 2sqrt(6)sqrt(2)
8=6+2
so 8+4sqrt3=(sqrt6+sqrt2)^2
cos a = (sqrt6+sqrt2)/4
we can calculate sin a by the pythagorean theorem
1-cos^2=sin^2
or by double angle identity
sin(2a)= 2 cos a sin a
sin 30º =2(sqrt6+sqrt2)*sin a/4
1/2 = sin a(sqrt6+sqrt2)/2
1= sin a (sqrt6+sqrt2)
multiply by |sqrt 6-sqrt2| on both sides
sqrt6-sqrt2=(6-2)sin a
so sin 15º = sqrt6-sqrt2/4
sin 15º>0 so sin 15º= |(sqrt6-sqrt2)/4|
sin 15º=(sqrt6-sqrt2)/4
AB= C*cos15º
AB=2sqrt2(sqrt6+sqrt2)/4=(sqrt2)*(sqrt6)+(sqrt2)^2/2=(sqrt12+2)/2=2*sqrt3+2/2=sqrt3+1
AC=C*sin 15º=2sqrt2(sqrt6-sqrt2)/4=(sqrt2)*(sqrt6)-(sqrt2)^2/2=(sqrt12-2)/2=2*sqrt3-2/2=sqrt3-1
This is very interesting, many thanks, Sir!
φ =30°; ∆ ABC → AC = c; BC = y; AB = x; xy/2 = 1 → y = 2/x; CAB = φ/2; sin(ABC) = 1; x = ? y = ? c = ?
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)(1 + cos(φ)) = (√2/4)(√3 + 1) →
tan(φ/2) = sin(φ/2)/cos(φ/2) = (1/2)(√3 - 1)^2 = y/x = 2/x^2 →
x = √3 + 1 → y = √3 - 1 → c = 2√2
Excellent!
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(1)^2=1 3x(5°)=15°x3x(5°)=15°x {15°x+15°x+90°}=120°x^2 {1+130°x^2}=131°x^2 {131°x^2-180°}=√49°x^2 7^√2 x^2 7^√2^1 x^2 7^√1^√1 x^272 (x+2x-7)
Let BC=a AB=b, and CA=c.
ba/2 = 1
ab = 2
tan(x-y) = (tan(x)-tan(y))/(1+tan(x)tan(y))
tan(45-30) = (tan(45)-tan(30))/(1+tan(45)tan(30))
tan(15) = (1-(1/√3))/(1+(1/√3))
tan(15) = ((√3-1)/√3)/((√3+1)/√3)
tan(15) = (√3-1)/(√3+1) = O/A
ab = 2
(√3-1)k•(√3+1)k = 2
k²(3-1) = 2
k² = 1
k = 1
a = √3 - 1
b = √3 + 1
BC² + AB² = CA²
(√3-1)² + (√3+1)² = c²
c² = 3 - 2√3 + 1 + 3 + 2√3 + 1
c = √8 = 2√2
AB = √3 + 1 cm
BC = √3 - 1 cm
CA = 2√2 cm
Thanks ❤️
Достроим еще один такой же треугольник, так, чтобы угол при вершине был 30. Используя формулу площади треугольника, найдем сторону AC. Площадь нового треугольника 2. sin(30)=0.5. По теореме косинусов найдем удвоенную сторону BC. И по теореме Пифагора AB.
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In this case, I went a very complicated way of first calculating the circle division as in the approximation of a circle area into 12 pieces based on a known radius, then used that with the area of a piece set to 1 (getting AC), then calculated the other two sides by the ratios gotten from the circle division calculation.
As it turns out, the resulting formulas look much more complicated (for example sqrt(4 - 2*sqrt(3)) for BC), but result in the exact same values (had to quadrate and simplify the term three times to prove that this is equal to sqrt(3) - 1).
The nice takeaway is, as always: "All roads lead to Rome" :D
And the nice side effect is: Have been learning a few things about the history of approximation of PI.
Sin15=(*3-1)/2.*2...(*=squre root
Cos15=(*3+1)/2.*2
Sin15=sin(45-30)=
Sin45.cos30+cos45.sin30{sin(-x)=-sinx and cos(-x)=cosx }
(1/*2).(*3/2)-(1/*2).(1/2)=
(*3/2.*2)-(1/2.*2)=
(*3-1)/2.*2
Applying the formula
Cos*x=1-sin^x, we shall get
Cos15=(*3+1)/2.*2
So,
Cot15=cos15/sin15
={(*3+1)/2.*2}/{(*3-1)/2*2}=
(*3+1)/(*3-1)=
(*3+1)(*3+1)/(*3-1)(*3+1)
=(4+2.*3)/(3-1)=
2(2+*3)/2=
2+*3=
b/p =2+*3.....Eqn1
According to the question
(1/2)bp=1=
bp=2.......eqn2
Eqn1 × Eqn2
(b/p)×bp=(2+*3).2
b^2=(4+2.*3)...(^=Squre)
b^2=(*3+1)^2
b=(*3+1)
Again
pb=2
P(*3+1)=2
P=2/(*3+1)
p={(*3+1)(*3-1)}/(*3+1)
=*3-1
H=*{(p^2)+(b^2)}
=*{(*3-1)^2+(*3+1)^2}
=*[2.{(*3)^2+1^2}]
=*{2(3+1)}
=*(2×4)
=*8
=2.*2
Sin(2x)=2sin(x)cos(x)
Sin(30)=1/2=2sin(15)cos(15)
1/4=sin(15)cos(15)
Area=r^2sin(15)cos(15)/2
Area=r^2(1/4)/2=r^2/8=1
r^2=8 and r=2sqrt(2)
AB=rcos(15)=2sqrt(2)cos(15)
BC=rsin(15)=2sqrt(2)sin(15)
This is not a very difficult Problem, but without a Calculator is not that simple. One must use Trigonometric Equality.
To solve this Geometrically one just have to solve the following System of Equations:
1) x * y = 2
2) y = tan(pi/12) * x
Solution : x ~ 2,732 and y ~ 0,732
I must work it out.
The 15°-75°-90° right triangle appears so frequently in problems that it is worthwhile to take note of the properties below. The 1) properties are derived in the video. The 2) properties can readily be derived from the formula for area = (1/2) (base) (height) and the properties of similar triangles.
1) The ratio of its 3 sides can be expressed as (√3 - 1):(√3 + 1):2√2. If the shorter side is given a length of 1, the longer side has length 2 + √3. If the longer side is given a length of 1, the shorter side has length 2 - √3. In both these cases, the expression for the hypotenuse will contain a radical within a radical.
2) If the hypotenuse has length h, the area of the triangle is h²/8 and, if the hypotenuse is treated as the base, the height is h/4.
In this problem, we are given an area of 1, so h²/8 = 1, h² = 8 and h = √8 = 2√2. We make use of the ratio of the 3 sides and find that, with the hypotenuse conveniently having a length of 2√2, the two side lengths are (√3 - 1) and (√3 + 1).
If you are taking a class on geometry and you are not given verbal permission in class, you can ask your teacher if it is OK to memorize these properties and use them on exams, without deriving them as part of the problem solution.
Thanks Sir
That’s very nice
Wonderful solution .
❤❤❤❤
You successfully solve it without any trigonometrical means, , but actually the simpler methods is to adopt the different angle angle formulas sin 60-45 and cos 60-45🙃🙂🙃
c cos 15=a, c sin 15=b, 1=1/2 a×b=1/2 c^2 sin 15 cos 15=1/4 c^2 sin 30=1/8 c^2, c^2=8, c=2sqrt(2), cos 15=sqrt((1+cos 30)/2)=sqrt((1+sqrt(3)/2)/2)=sqrt((4+2sqrt(3))/8)=(sqrt(3)+1)/(2sqrt(2)), sin 15=sqrt((1-sqrt(3)/2)/2)=sqrt((4-2sqrt(3))/8)=(sqrt(3)-1)/(2sqrt(2)), so a=sqrt(3)+1, b=sqrt(3)-1.😊😊😊😊😊😊😊
Excellent!
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If AC=a, then we use sin(15) to know BC and AB, then ABxBC=2, so we know a, so we can know AC, AB, BC. It is much easier.
Это фигура часть квадрата с построенной внутри равносторонним треугольником со стороной равной стороне квадрата. А дальше кроме теоремы Пифагора ничего знать не надо. Сторона квадрата равен 2Х.
А У=2Х-Н где Н высота равностороннего треугольника со стороной 2Х. Н= Х√3
Получается У= Х(2-√3)
ХУ= 2=Х^2(2-√3).... А дальше всё тоже самое. Спасибо
Very easy... First AB.BC =area.2 = 2. Then BC = AB.tan(15°) = AB.(2 - sqrt(3)), and so 2 = AB. AB.(2 -sqrt(3)),
which gives that AB^2 = 2/(2 -sqrt(3)) = 2.(2 +sqrt(3)) = 1 + 3 + 2.sqrt(3) = (1 + sqrt(3))^2. Finally we have: AB = 1 +sqrt(3)
Now BC = 2/AB = 2/(1 +sqrt(3)) = sqrt(3) - 1. And the Pythagorean theorem gives then that AC^2 = AB^2 + BC^2 = (1 + 3 +2.sqrt(3)) + (1 + 3 - 2.sqrt(3)) = 8
Then AC = sqrt(8) = 2.sqrt(2). Conclusion: AB = sqrt(3) - 1; BC = sqrt(3) - 1; AC = 2.sqrt(2).
Excellent!
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I have to say … I worked on this for about an hour, and got something WAY more complicated, but which turned out to deliver the same numbers!
Neat trick multiplying by conjugates. Forgot that part. Good one.
Basically, I took a 'whole copy' of the triangle and flipped it over connecting on the hypotenuses.
The total trapezoid allowed the tan 15° to be found as 1/(2+√3).
From there, we have a ratio between the base length and the height of the original triangle, and need to scale it.
Thanks AGAIN!
⋅-=≡ GoatGuy ✓ ≡=-⋅
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√3-1; √3+1; 2√2
1/ Label BC=a and AB= b, (a we have the triangle ABD is a right isosceles.
From C drop the height CH to AD---> the triangle ACH is a special 30-60-90 one and the triangle CHD is a 45-90-45 one.
We have CD= b-a----> CH= (b-a)/sqrt2.
Moreover CH= AC/2----> CH= c/2 -----> (b-a)/sqrt2= c/2----> b-a= c/sqrt2---> sq(b-a)= (sq c)/2----> sq b+sqa -2 ba =sq c/2
because a.b= 2----> sq c -4 = sqc/2---> 2sqc-8 = sq c-----> sq c= 8-----> c = 2sqrt 2
2/ Calculating a and b
we have a.b= 2 and b-a = c/sqrt2 =2---> b= a+2 ---> a.(a+2)= 2---> sqa + 2a -2 =0----> a =sqrt3-1 (negative answer rejected)
and b = sqrt3 +1
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What was that?!? LOL Wow!!! Amazing.
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AB=2...b/a=tg15=√(1-√3/2)/(1+√3/2)=√(2-√3)/(2+√3)=(2-√3)/1=2-√3...b=(2-√3)a ..(2-√3)a^2=2..a^2=2/(2-√3)=2(2+√3)=4+2√3=(1+√3)^2...a=1+√3..b=(2-√3)(1+√3)=-1+√3...c^2=(1+√3)^2+(√3-1)^2=4+2√3+4-2√3=8...c=2√2
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Very good teacher!!
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a=0.73 ; b=2.73; c=2.83 ❤❤❤thanks sir.
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According to trigonometric tables: sin 15° = ¼ (√6 − √2) = (√3 − 1) / (2√2) and cos 15° = ¼ (√6 + √2) = (√3 + 1) / (2√2).
Thus, according to SOH-CAH-TOA: the hypotenuse is 2√2, the opposite side is √3 − 1 and the adjacent side is (√3 + 1).
But I have a feeling that if using a calculator was forbidden, the good old trigonometric tables where forbidden too 🤔
😂
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Easiest way .
Let hypotenuse be r.
Then x= r cos 15. y = r sin 15
area = 1 = r cos 15 *r sin 15
=. r *r 2*sin 15* cos 15
Solving r=2 * sqrt 2
Also x= r cos 15
r* cos (45 -30)
y = r* sin(45- 30)
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I like the questions brought by you.
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Thank you!
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it can be solved by using the identity tan(A-B) and then applying pythagoras theorom.
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В прямоугольном треугольнике высота проведенная из вершины прямого угла равна 1/4 гипотенузы.Значит S=c2/8=1,c2=8..
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✨Magic! ✨
I liked it.