Why all these steps? If the area of rectangle is 2,the area of triangle will be 1. So we take tan 15°=width/length and taking the formula of area of triangle A=1/2*width*length. After substitution of the value,the value of lenth and width will be 2.8 and 0.7 units respectively.Thanks🙏
The 15°-75°-90° triangle, while not considered "special" in geometry, appears frequently enough in problems that it is worthwhile to have its properties handy. The ratio of sides (short):(long):(hypotenuse) can be written as (√3 − 1):(√3 + 1):(2√2). The area is hypotenuse squared divided by 8. Let the sides be (√3 − 1)m, (√3 + 1)m and hypotenuse (2√2)m. The area of each triangle in the rectangle is half the area of the rectangle, or 2/2 = 1. So, ((2√2)m)²/8 = 1, 8m²/8 = 1, m² = 1 and m = 1. Width = short side = (√3 − 1)m = (√3 − 1). Length = long side = (√3 + 1)m = (√3 + 1), as PreMath also found. Note that PreMath effectively derived the properties of the 15°-75°-90° triangle, except for the above formula for its area.
In case of any 🔺 formed by the diagonal width /length =tan 15 =(√3-1)/(√3+1) Now (√3-1)(√3+1)=2 This result matches with the given area of rectangle Hence width and length are √3 -1 and √3 +1 respectively.
There are two "Perfect Square Trinomials" and also "The Difference of Squares". So @ 7:45 that "Famous Identity" one can see the pattern visually if a is the side length of one Big happy square then (a-b)² is a happy square in the Big square and b² is another happy square in the Big square. There are two happy Rectangles b(a-b) in the Big happy square. ... so with some fancy algebraic nipplation we get: (a-b)²=a²-2b(a-b)-b² =a²-2ab+2b²-b² =a²-2ab+b² ... @ 8:35 hopefully you've had enough coffee too start tweakin to get y² into a manageable form to apply that Famous "Difference of Squares" Identity! Good luck! 😊
Let's note a = AB and b = AD. We have a.b = 2, and in triangle ABC b/a = tan(15°) = 2 -sqrt(3), so 2 = (2 - sqrt(3)).(a^2), which gives that a^2 = 2/(2 -sqrt(3)) or a^2 = 4 + 2.sqrt(3) = (1 + sqrt(3))^2. we finally have that a = 1 + sqrt(3) and b = 2/(1 +sqrt(3)) = -1 + sqrt(3). (Very easy if tan(15°) is known.)
Answer 0.732051 and 2.732051 or sqrt 3 - 1 and sqrt 3 -1 note that in a 75, 15, 9and 0 right triangle, the ratio of the three sites is sqrt 3- 1 sqrt 3 + 1 and the hypotenuse is 2 sqrt 2 Since the area of a triangle is L * W, then sqrt 3 -1 * sqrt 3 + 1 --------------- 3 -1 = 2 Since the area is given as 2, then the answer is length sqrt 3 + 1 and width sqrt 3 -1
A = lw lw = 2 Label AB = l & BC = w. By definition of acute angles in a right triangle, m∠ACB = 75°. Draw a segment thru point C and a point E on side AB, such that m∠BCE = 60°. So, m∠ACE = 75°- 60° = 15°. As a result, △AEC is an isosceles triangle and △CBE is a special 30°-60°-90° right triangle. So, AE = CE. Use the properties of 30°-60°-90° right triangles. b = a√3 BE = w√3 c = 2a CE = 2w So, AE = 2w. By the Segment Addition Postulate, AB = l = AE + BE = 2w + w√3 = w(2 + √3). lw = 2 w * w * (2 + √3) = 2 w² * (2 + √3) = 2 w² = 2/(2 + √3) = [2(2 - √3)]/(4 - 3) = (4 - 2√3)/1 = 4 - 2√3 = (3 + 1) - 2√3 = (√3)² + 1² - 2√3 = (√3 - 1)² w = √3 - 1 Remember the width can't be negative. l = w(2 + √3) = (-1 + √3)(2 + √3) = -2 + 2√3 - √3 + 3 = 1 + √3 So, the side lengths of the rectangle are as follows: Length = 1 + √3 u ≈ 2.73 u Width = -1 + √3 u ≈ 0.73 u
To avoid the 15,75,90 degree right triangle we can cut the rectangle in 2 parts along the diagonal AC and then we can draw an isosceles triangle with the same area of the rectangle and with the upper angle A = 30° (being the two angle at the base = 75°). The area of such isosceles triangle is: Area = 1/2*x*x*sin 30° = 2 ( x are the two equal sides) x² = 8 x = 2√ 2 that is the measure of diagonal AC (to avoid trigonometry you can draw a right triangle of 30,60,90 degree inside the isosceles triangle, you will find that area = 1/2*x*2x=2 and AC = 2x) Finally we can solve the problem knowing that in the right triangle ABC in the video, its area = 1 and AC = 2√ 2 1/2*AB*BC = 1 (area) AB² + BC² = AC² so, setting AB=a and BC=b we have the following system: a*b = 2 a² + b² = ( 2√ 2)² = 8 (a+b)² = a² + b² + 2ab = 8 + 2*2 = 12 a + b = 2√ 3 we can find a and b with : t² - (a+b)t + (a*b) = 0 t² - 2√ 3t + 2 = 0 a = √ 3 + 1 b = √ 3 - 1
@@PreMath It's nice to have the practice. Algebra and trigonometry have served me well over the years, going back to slide rules. Funny, in my friends circles, very few use them.
Here's another trigonometry approach avoiding approximations: let d be the long diagonal. We have y = sin(15°)d, x = cos(15°)d So 2 = xy = sin(15°)cos(15°)d² (1) Using the sin double-angle trig identity we see sin(15°)cos(15°) = sin(30°)/2 = .5/2 = .25 so from (1) we have d² = 8 (2) By Pythagoras, d² = x² + y² so combining with (1) and (2) x² + y² + 2xy = 8 + 4 = 12 which gives (x + y)² = 12 x + y = 2√3 y = 2√3 - x (3) Substituting back into (1) x(2√3 - x) = 2 which gives the quadratic: x² - x(2√3) + 2 = 0 easily solved by the quadratic formula to give x = √3 + 1 and the short side is the other solution y = √3 - 1
In case of any of the two triangles formed by the diagonal Width /length =tan 15 =(√3-1)/(√3+1) Now width is (√3-1)x & length is (√3+1)x Now Width *length =2x^2 Hence 2x^2=2 >x =1 Then width is √3 -1 & length is √3 +1
Let AB = CD = x and BC = DA = y. The area of the rectangle will be xy = 2. Let ∠CAB = θ. tanθ = BC/AB tan15° = y/x (√3-1)/(√3+1) = y/x (√3+1)y = (√3-1)x (√3+1)(2/x) = (√3-1)x 2(√3+1) = (√3-1)x² x² = 2√3+1)/(√3-1) x² = 2(√3+1)(√3+1)/(√3-1)(√3+1) x² = 2(√3+1)²/(3-1) = (√3+1)² [ x = √3 + 1 ≈ 2.732 units ] y = 2/x = 2/(√3+1) y = 2(√3-1)/(√3+1)(√3-1) y = 2(√3-1)/(3-1) [ y = √3 - 1 ≈ 0.732 units ]
another approach The area of a 15, 75, 90-degree right triangle can be determined by hypotenuse squared /8. Hence, h^2/8 = a (partly due to sin 75. sin 15= 1/4 and sin 90 =1 and the law of sines) Hence, 1= h^2/8 (Since if the area of the rectangle = 2, then the area of the triangle is one-half or 1) h^2=8 h= sqrt 8 Let the width of the rectangle = a and the length b, then ab=2 and a^2 + b^2 = (sqrt 8)^2 a^2 + b^2 = 8 Since ab=2 , then a=2/b (2/b)^2 + b^2 =8 4/b^2 + b^2 = 8 4 + b^4 = 8b^2 Let n= b^2 Hence, n^2= b^4 Hence 4 + n^2= 8n n^2-8n+ 4 n= 0.535898 and n= 7.4641 Hence a^2 = 0.535898 and b^2= 7.4641 hence a = sqrt 0.535898 or 0.7320505. Answer Hence, b = sqrt 7.4641 or 2.7320505 Answer Proof why the area of a 15-75-90 right triangle is equal hypotenuse squared/8 So why h^2/8 = area? note a =BC b=AB Law of Sines: a/sin A = b/sin B = c/sin C Any two of the variables can be used So a/sin A = c /sin C and b/sin B = c/sin C Hence, a/sin 15 = c/sin 90 Hence, a = c*sin 15 / sin 90 and b = c* sin 75/ sin 90 Recall the area of a triangle = 1/2 base* height Hence, ab/2 Hence, (c*sin15/sin90)(c*sin75/sin90)* 1/2 (This is a*b/2) Let's rearrange the above c*c* sin15* sin75* sin15* sin75*1/sin 90*1/sin90* 1 c^2 *1/4*1/4 / 1/1 * 2 c^2 * 1/8 c^2/8 but c^2= hypotenuse square Hence, the area of a 75-90-15 right triangle can be determined if only the hypotenuse is given, and it is the hypotenuse squared divided by 8 or c^2/8 or h^2/8 (I used h for the hypotenuse above)
Thanks I love it. in a right triangle with a 15° angle, the altitude to the hypotenuse is one fourth of the hypotenuse. So AC = 2 rad.2. And from xy=2 we find x and y
Solution: tan 15° = h/b h = b . tan 15° ... ¹ Area = 2 b h = 2 ... ² Replacing ¹ in ² b. (b . tan 15°) = 2 b² = 2/tan 15° b ≈ 2.7321 Units ✅ 2.7321 . h = 2 h ≈ 0.7321 Units ✅
My way of solution ▶ In this rectangle ABCD [AB]= a [BC]= b Let’s consider the right triangle ΔABC. We know that: tan(15°)= [BC]/[AB] 0,2679= b/a ⇒ b= 0,2679 a a*b = 2 b= 0,2679 a ⇒ a*0,2679 a= 2 0,2679 a²= 2 a²= 7,4641 a= 2,732 b= 0,2679 a b= 0,732 ⇒ Thus, we have : a= 2,732 length units b= 0,732 length units
Another super and well explained demonstration ! Thanks for sharing Professor . I came across this question ,……..the average age of Men and Women in a group is 25 yrs . ….if the ave age of the Men is 27 and, the ave age of the Women is 20 What is the percentage of : Males , Females ? I calculated thus …: 27m +20w = 25m + 25w ……2m=5w ……..ratio 2:5 ….2 men for every 5 women this equates to 2/7 men and 5/7 women 200/7 = 28.571429% men 500/7 = 71.428571% women . Although the question was only to ascertain the percentage in this group that were men and what percentage were women ….. Is it possible with the available information to calculate the number of men and number of women ? ……is there an equation to calculate this enigma ? Thanks
Your first part answer is correct! With given info, it's not possible to calculate the exact number of men and women! Thanks for sharing the interesting enigma😀
I just "cheated" and looked up *tan(15) = 2 - sqrt(3)* as I remembered it was something like that and didn't want to convert to decimals. So 2 = xy = (x)(x*(2-sqrt(3)) = x^2 * (2-sqrt(3)), then pretty much followed the rest of the steps to find x, then y.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) AB = X 02) BC = Y 03) X * Y = 2 04) tan(15º) = Y / X 05) tan(15º) ~ 0,268 06) 0,268 ~ Y / X 07) System of Equations : a) Y / X ~ 0,268 ; Y ~ 0,268X b) X * Y ~ 2 ; X * (0,268X) = 2 ; 0,268X^2 = 2 ; X^2 = 7,5 ; X = 2,732 08) Y = 0,268 * 2,732 ; Y = 0,732 Therefore, OUR BEST ANSWER : Length AB ~ 2,732 Linear Units, and Length BC ~ 0,732 Linear Units
Thank you! The quest to find a special triangle, solving the problem.
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a bit of engineering:
1. let w= width, l=length:
2. {l*w=2; w/l=tan(D15)};
3. w= l*tan(15);
4. tan(15) = 2-sqrt(3);
5. l*l*tan(15)=2;
l= sqrt(2/tan(15)) = sqrt(3)+1
6. w = (1+sqrt(3)) *tan(15)= sqrt(3)-1.
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Why all these steps? If the area of rectangle is 2,the area of triangle will be 1. So we take tan 15°=width/length and taking the formula of area of triangle A=1/2*width*length. After substitution of the value,the value of lenth and width will be 2.8 and 0.7 units respectively.Thanks🙏
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I was shocked by watching the video length.
I think it might be explained without using trigonometry, but it was used.
You can do tan 15 = tan (45-30)
The 15°-75°-90° triangle, while not considered "special" in geometry, appears frequently enough in problems that it is worthwhile to have its properties handy. The ratio of sides (short):(long):(hypotenuse) can be written as (√3 − 1):(√3 + 1):(2√2). The area is hypotenuse squared divided by 8. Let the sides be (√3 − 1)m, (√3 + 1)m and hypotenuse (2√2)m. The area of each triangle in the rectangle is half the area of the rectangle, or 2/2 = 1. So, ((2√2)m)²/8 = 1, 8m²/8 = 1, m² = 1 and m = 1. Width = short side = (√3 − 1)m = (√3 − 1). Length = long side = (√3 + 1)m = (√3 + 1), as PreMath also found.
Note that PreMath effectively derived the properties of the 15°-75°-90° triangle, except for the above formula for its area.
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@@PreMath and thank you for the compliment! Please keep up the good work posting challenging geometry problems with solutions!
In case of any 🔺 formed by the diagonal
width /length =tan 15
=(√3-1)/(√3+1)
Now
(√3-1)(√3+1)=2
This result matches with the given area of rectangle
Hence width and length are √3 -1 and √3 +1 respectively.
A = b.h = 2 cm² --> h=A/b
h= b tan15°
Equalling:
A/b = b tan15°
b² = A/tab15° = 2/tan15°
b = 2,73cm = √3+1 cm
h = A/b= 0,73cm= √3-1 cm
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There are two "Perfect Square Trinomials" and also "The Difference of Squares". So @ 7:45 that "Famous Identity" one can see the pattern visually if a is the side length of one Big happy square then (a-b)² is a happy square in the Big square and b² is another happy square in the Big square. There are two happy Rectangles b(a-b) in the Big happy square. ... so with some fancy algebraic nipplation we get:
(a-b)²=a²-2b(a-b)-b²
=a²-2ab+2b²-b²
=a²-2ab+b²
... @ 8:35 hopefully you've had enough coffee too start tweakin to get y² into a manageable form to apply that Famous "Difference of Squares" Identity! Good luck! 😊
😀
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Let's note a = AB and b = AD. We have a.b = 2, and in triangle ABC b/a = tan(15°) = 2 -sqrt(3), so 2 = (2 - sqrt(3)).(a^2), which gives that a^2 = 2/(2 -sqrt(3))
or a^2 = 4 + 2.sqrt(3) = (1 + sqrt(3))^2. we finally have that a = 1 + sqrt(3) and b = 2/(1 +sqrt(3)) = -1 + sqrt(3). (Very easy if tan(15°) is known.)
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Answer 0.732051 and 2.732051 or sqrt 3 - 1 and sqrt 3 -1
note that in a 75, 15, 9and 0 right triangle, the ratio of the three sites is sqrt 3- 1 sqrt 3 + 1
and the hypotenuse is 2 sqrt 2
Since the area of a triangle is L * W, then sqrt 3 -1
* sqrt 3 + 1
---------------
3 -1 = 2
Since the area is given as 2, then the answer is length sqrt 3 + 1 and width sqrt 3 -1
Nice! φ = 30°; ∎ABCD → AB = CD = a; AD = BC = b; DAB = ABC = 3φ
CAB = φ/2; ab/2 = 2 → b = 4/a
tan(φ/2) = 2 - √3 = b/a → b = a(2 - √3) = 4/a → a = (√2)(√3 + 1) → b = (√2)(√3 - 1)
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Thanks from Morocco
شكرا لكم
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A = lw
lw = 2
Label AB = l & BC = w.
By definition of acute angles in a right triangle, m∠ACB = 75°.
Draw a segment thru point C and a point E on side AB, such that m∠BCE = 60°.
So, m∠ACE = 75°- 60° = 15°.
As a result, △AEC is an isosceles triangle and △CBE is a special 30°-60°-90° right triangle.
So, AE = CE. Use the properties of 30°-60°-90° right triangles.
b = a√3
BE = w√3
c = 2a
CE = 2w
So, AE = 2w. By the Segment Addition Postulate, AB = l = AE + BE = 2w + w√3 = w(2 + √3).
lw = 2
w * w * (2 + √3) = 2
w² * (2 + √3) = 2
w² = 2/(2 + √3)
= [2(2 - √3)]/(4 - 3)
= (4 - 2√3)/1
= 4 - 2√3
= (3 + 1) - 2√3
= (√3)² + 1² - 2√3
= (√3 - 1)²
w = √3 - 1
Remember the width can't be negative.
l = w(2 + √3)
= (-1 + √3)(2 + √3)
= -2 + 2√3 - √3 + 3
= 1 + √3
So, the side lengths of the rectangle are as follows:
Length = 1 + √3 u ≈ 2.73 u
Width = -1 + √3 u ≈ 0.73 u
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Solution:
a = horizontal side of the rectangle,
b = vertical side of the rectangle.
(1) a*b = 2
(2) b/a = tan(15°) |*a ⟹ (2a) b = a*tan(15°) |in (1) ⟹
(1a) a*a*tan(15°) = a²*tan(15°) = 2 |/tan(15°) ⟹
(1b) a² = 2/tan(15°) |√() ⟹
(1c) a = √[2/tan(15°)] ≈ 2.7321 |in (2a)
(2b) b = √[2/tan(15°)]*tan(15°) = √[2*tan(15°)] ≈ 0.7321
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Nice video sir
🙏
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Tan75° = 2 + √3
width = x, length = (2 + √3)x
rectangle area = (2 + √3)x^2 = 2
width = x = √(4 - 2√3) = √3 - 1
length = (2 + √3)(√3 - 1) = 1 + √3
To avoid the 15,75,90 degree right triangle we can cut the rectangle in 2 parts along the diagonal AC and then we can draw an isosceles triangle with the same area of the rectangle and with the upper angle A = 30° (being the two angle at the base = 75°). The area of such isosceles triangle is:
Area = 1/2*x*x*sin 30° = 2 ( x are the two equal sides)
x² = 8
x = 2√ 2 that is the measure of diagonal AC
(to avoid trigonometry you can draw a right triangle of 30,60,90 degree inside the isosceles triangle, you will find that area = 1/2*x*2x=2 and AC = 2x)
Finally we can solve the problem knowing that in the right triangle ABC in the video, its area = 1 and AC = 2√ 2
1/2*AB*BC = 1 (area)
AB² + BC² = AC²
so, setting AB=a and BC=b we have the following system:
a*b = 2
a² + b² = ( 2√ 2)² = 8
(a+b)² = a² + b² + 2ab = 8 + 2*2 = 12
a + b = 2√ 3
we can find a and b with : t² - (a+b)t + (a*b) = 0
t² - 2√ 3t + 2 = 0
a = √ 3 + 1
b = √ 3 - 1
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Bom dia Mestre
Hello dear❤️
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√3+1; √3-1
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tan 15 ° = 0.268
ab = 2
a(0.268a) = 2
a^2 = 2/0.268
a = 2.732
b = 0.732
My way exactly. This is how an engineer would state the solution. Pre-Math states it in the form of a mathematician.
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@@PreMath It's nice to have the practice.
Algebra and trigonometry have served me well over the years, going back to slide rules.
Funny, in my friends circles, very few use them.
@@calvinmasters6159
Super!
Many thanks
Here's another trigonometry approach avoiding approximations:
let d be the long diagonal. We have y = sin(15°)d, x = cos(15°)d
So 2 = xy = sin(15°)cos(15°)d² (1)
Using the sin double-angle trig identity we see
sin(15°)cos(15°) = sin(30°)/2 = .5/2 = .25
so from (1) we have d² = 8 (2)
By Pythagoras, d² = x² + y² so combining with (1) and (2)
x² + y² + 2xy = 8 + 4 = 12 which gives
(x + y)² = 12
x + y = 2√3
y = 2√3 - x (3)
Substituting back into (1)
x(2√3 - x) = 2 which gives the quadratic:
x² - x(2√3) + 2 = 0 easily solved by the quadratic formula to give
x = √3 + 1 and the short side is the other solution
y = √3 - 1
In case of any of the two triangles formed by the diagonal
Width /length =tan 15
=(√3-1)/(√3+1)
Now width is (√3-1)x
& length is (√3+1)x
Now
Width *length
=2x^2
Hence 2x^2=2
>x =1
Then width is √3 -1
& length is √3 +1
Let AB = CD = x and BC = DA = y. The area of the rectangle will be xy = 2. Let ∠CAB = θ.
tanθ = BC/AB
tan15° = y/x
(√3-1)/(√3+1) = y/x
(√3+1)y = (√3-1)x
(√3+1)(2/x) = (√3-1)x
2(√3+1) = (√3-1)x²
x² = 2√3+1)/(√3-1)
x² = 2(√3+1)(√3+1)/(√3-1)(√3+1)
x² = 2(√3+1)²/(3-1) = (√3+1)²
[ x = √3 + 1 ≈ 2.732 units ]
y = 2/x = 2/(√3+1)
y = 2(√3-1)/(√3+1)(√3-1)
y = 2(√3-1)/(3-1)
[ y = √3 - 1 ≈ 0.732 units ]
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Ótima saída para quem não sabe encontrar a tangente de 15°! Parabéns! 🎉🎉🎉
Fico feliz em ouvir isso! 😀
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Obrigado pelo feedback ❤️
A = b.h = 2 cm²
tan15°= h/b --> h= b tan15°
Replacing :
A = b² tan15° = 2 cm²
b² = 2/tan15°
b = 2,73cm = √3+1 cm
h = A/b= 0,73cm = √3-1 cm
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My approach as well
Let's face this challenge:
.
..
...
....
.....
Let a=AB=CD and let b=BC=DA be the side lengths of the rectangle. Since ABC is a right triangle, we obtain the following two equations for our two unknown quantities a and b:
A(ABCD) = a*b
b/a = tan(∠BAC) = tan(15°)
tan(15°)
= tan(45° − 30°)
= [tan(45°) − tan(30°)]/[1 + tan(45°)*tan(30°)]
= (1 − 1/√3)/(1 + 1/√3)
= (√3 − 1)/(√3 + 1)
2 = a*b ∧ b = a*tan(15°) = a*(√3 − 1)/(√3 + 1) ⇒ 2 = a²*(√3 − 1)/(√3 + 1)
2 = a²*(√3 − 1)/(√3 + 1)
2*(√3 + 1) = a²*(√3 − 1)
2*(√3 + 1)² = a²*(√3 − 1)*(√3 + 1)
2*(√3 + 1)² = a²*(3 − 1)
2*(√3 + 1)² = a²*2
(√3 + 1)² = a²
⇒ a = √3 + 1
⇒ b = (√3 + 1)*(√3 − 1)/(√3 + 1) = √3 − 1
Best regards from Germany
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another approach
The area of a 15, 75, 90-degree right triangle can be determined by
hypotenuse squared /8. Hence, h^2/8 = a (partly due to sin 75. sin 15= 1/4 and
sin 90 =1 and the law of sines)
Hence, 1= h^2/8
(Since if the area of the rectangle = 2, then the area of the triangle is one-half or 1)
h^2=8
h= sqrt 8
Let the width of the rectangle = a and the length b,
then ab=2 and
a^2 + b^2 = (sqrt 8)^2
a^2 + b^2 = 8
Since ab=2 , then a=2/b
(2/b)^2 + b^2 =8
4/b^2 + b^2 = 8
4 + b^4 = 8b^2
Let n= b^2
Hence, n^2= b^4
Hence 4 + n^2= 8n
n^2-8n+ 4
n= 0.535898 and n= 7.4641
Hence a^2 = 0.535898 and b^2= 7.4641
hence a = sqrt 0.535898 or 0.7320505. Answer
Hence, b = sqrt 7.4641 or 2.7320505 Answer
Proof why the area of a 15-75-90 right triangle is equal hypotenuse squared/8
So why h^2/8 = area?
note a =BC b=AB
Law of Sines:
a/sin A = b/sin B = c/sin C
Any two of the variables can be used
So a/sin A = c /sin C
and b/sin B = c/sin C
Hence, a/sin 15 = c/sin 90
Hence, a = c*sin 15 / sin 90
and b = c* sin 75/ sin 90
Recall the area of a triangle = 1/2 base* height
Hence, ab/2
Hence, (c*sin15/sin90)(c*sin75/sin90)* 1/2 (This is a*b/2)
Let's rearrange the above
c*c* sin15* sin75* sin15* sin75*1/sin 90*1/sin90* 1
c^2 *1/4*1/4 / 1/1 * 2
c^2 * 1/8
c^2/8
but c^2= hypotenuse square
Hence, the area of a 75-90-15 right triangle can be determined if only the hypotenuse
is given, and it is the hypotenuse squared divided by 8 or c^2/8 or h^2/8 (I used
h for the hypotenuse above)
Length =*3+1, breadth =*3-1 (May be )
*=read as square root
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x^2 * tan(15) = 2
x = sqrt(2/tan(15)) = 2.73205
y = x * tan(15) = 0.73205
tan 15°=2-√3
2-√3=a/b
ab=2
b=2/a
2-√3=a/2/a
2-√3=a² /2
a²=2(2- √3)
a²=4-2√3
a²=(3+1-2√3)
a=√3-1
2=(√3-1)b
b=2/(√3-1)
b=√3+1
a=√3-1
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Length = L, width = L.Tan(15)
2 = L².Tan(15)
L=√(2/Tan(15) )
2.732 and 0.732
Thanks I love it. in a right triangle with a 15° angle, the altitude to the hypotenuse is one fourth of the hypotenuse. So AC = 2 rad.2. And from xy=2 we find x and y
note diagonal.is 2ζ2
Solution:
tan 15° = h/b
h = b . tan 15° ... ¹
Area = 2
b h = 2 ... ²
Replacing ¹ in ²
b. (b . tan 15°) = 2
b² = 2/tan 15°
b ≈ 2.7321 Units ✅
2.7321 . h = 2
h ≈ 0.7321 Units ✅
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l= √3+1, b=√3-1
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My way of solution ▶
In this rectangle ABCD
[AB]= a
[BC]= b
Let’s consider the right triangle ΔABC.
We know that:
tan(15°)= [BC]/[AB]
0,2679= b/a
⇒
b= 0,2679 a
a*b = 2
b= 0,2679 a
⇒
a*0,2679 a= 2
0,2679 a²= 2
a²= 7,4641
a= 2,732
b= 0,2679 a
b= 0,732
⇒
Thus, we have :
a= 2,732 length units
b= 0,732 length units
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xy=2,y/x=2/x^2
tan15°=tan(45°-30°)=(√3-1)/(√3+1)=2/(√3+1)^2=y/x=2/x^2
x^2=(√3+1)^2,x=√3-1,y=2/(3+1)=(√3-1)(√3+1)/(√3+1)=√3-1
x=√3+1. ,y=√3-1
How can possible if we divide a rectangle than one angle 15 degree.
Another super and well explained demonstration ! Thanks for sharing Professor .
I came across this question ,……..the average age of Men and Women in a group is 25 yrs .
….if the ave age of the Men is 27 and, the ave age of the Women is 20
What is the percentage of : Males , Females ?
I calculated thus …: 27m +20w = 25m + 25w ……2m=5w ……..ratio 2:5 ….2 men for every 5 women this equates to 2/7 men and 5/7 women
200/7 = 28.571429% men
500/7 = 71.428571% women .
Although the question was only to ascertain the percentage in this group that were men and what percentage were women …..
Is it possible with the available information to calculate the number of men and number of women ? ……is there an equation to calculate this enigma ?
Thanks
Your first part answer is correct!
With given info, it's not possible to calculate the exact number of men and women!
Thanks for sharing the interesting enigma😀
@@PreMath …..Thanks Professor 🙏😉
Hard to see how a rectangle (ABCD) with an agree-upon area of 2 units, somehow becomes a new rectangle with 4.7229 units.
Sqrt6,sqrt(2/3).
I just "cheated" and looked up *tan(15) = 2 - sqrt(3)* as I remembered it was something like that and didn't want to convert to decimals. So 2 = xy = (x)(x*(2-sqrt(3)) = x^2 * (2-sqrt(3)), then pretty much followed the rest of the steps to find x, then y.
Now I’m trying to find tan(40) / tan(50) with drawing auxiliary lines.
90-15=75÷15=5)(0×0=0×5=0 area
Thanks for the feedback ❤️
Is E the midpoint?
No it is a point picked to give us gives us (30° at E and 60° at C) to use the 30° 60° 90° special triangle rules . Check it @4:13
@@Abby-hi4sf Makes sense.
I have method without trigonometry
Thanks for the feedback ❤️
We know,
breadth = length × tan(15°)
Length × breadth = 2
Length = √(2 ÷ tan(15°))
And breadth = √(2 × tan(15°))
Thank you very much.😊😊
STEP-BY-STEP RESOLUTION PROPOSAL :
01) AB = X
02) BC = Y
03) X * Y = 2
04) tan(15º) = Y / X
05) tan(15º) ~ 0,268
06) 0,268 ~ Y / X
07) System of Equations :
a) Y / X ~ 0,268 ; Y ~ 0,268X
b) X * Y ~ 2 ; X * (0,268X) = 2 ; 0,268X^2 = 2 ; X^2 = 7,5 ; X = 2,732
08) Y = 0,268 * 2,732 ; Y = 0,732
Therefore,
OUR BEST ANSWER :
Length AB ~ 2,732 Linear Units, and
Length BC ~ 0,732 Linear Units
Excellent!
Thanks for sharing ❤️