Let the hypotenuse have length c and the sides a and b. We can simplify the algebra by factoring the hypotenuse length by 23, leaving c = 4 and the area by (23)², leaving (1/2)ab = 2 or ab = 4. The resulting triangle is similar, so has the same corresponding angles. Now, we take an educated guess that the triangle is either a special 30°-60°-90° or 45°-45°-90° right triangle. We try 30°-60°-90°. The sides are 1, √3, 2 which we multiply by 2 to make the hypotenuse c = 4, so 2, 2√3, 4. The product ab is (2)(2√3) = 4√3, which is not the required 4. We try 45°-45°-90°. The sides are √2/2, √2/2, 1 which scales up to 2√2, 2√2, 4. The product ab = (2√2)(2√2) = 8, which is not 4. So, we take another educated guess that the triangle is 15°-75°-90°, which has sides (√3 - 1), (√3 + 1), 2√2, and we scale up to (√2)(√3 - 1), (√2)(√3 + 1), 4. The product ab = (√2)(√3 - 1)(√2)(√3 + 1) = (2)(√3 - 1)(√3 + 1) = (2)(3 - 1) = 4. We are in luck! So, this is a 15°-75°-90° triangle and x = 15° or 75°. Note that the 15°-75°-90° triangle is not designated as a "special" right triangle in geometry, while the 30°-60°-90° or 45°-45°-90° right triangles are. However, it appears frequently in problems, so it is worthwhile to know its side ratios.
ab=2116, a^2+b^2=92^2=8464, (a+b)^2=8464+2×2116=(46sqrt(6))^2, a, b are given from (46sqrt(6)+-46sqrt(2))/2=23sqrt(2)(sqrt(3)+-1), thus b/a=(sqrt(3)-1)/(sqrt(3)+1)=(1-1/sqrt(3)))/(1+1/sqrt(3))=tan(45-30)=tan(15), therefore x=15 or 90-15=75. 1/2×(sqrt(3^2=1/2×(4-2sqrt(3))=2-sqrt(3), (1-1/sqrt(3))/(1+1/sqrt(3))
Let BC = a, AB = b and AC = c. Then we have ½ ab = 1058 and a² + b² = 92². So a = 46√(2 − √3) and b = 46√(2 − √3) (2 + √3). Then a / b = 1 / (2 + √3). So we can conclude that the angle x is tan⁻¹ (1 / (2 + √3)) = 15°.
I propose something interesting (I hope so!): Let'a name a = BC and b = AC. We have: a^2 + b^2 = 92^2 = 8464 (Pythagorean theorem in triangle ABC) and a.b = 2.area of triangle ABC = 2116. Let's build a second degree equation whose solutions are a/b and b/a: This equation is (x -a/b).(x -b/a) = 0 or x^2 - (a/b + b/a).x + (a/b).(b/a) =0 So it is: x^2 - ((a^2 + b^2)/(a.b)).x+ 1 = 0 or x^2 - (8464/2116).x +1 = 0 or finally x^2 -4.x +1 = 0. Its solutions are 2 +sqrt(5) and 2 -sqrt(5) Conclusion: tan(x) = a/b = 2 - sqrt(5) and x = 15° or tan(x) = a/b = 2 +sqrt(5) = cotan(15°)n = tan (90° - 15°) = tan(75°) and x = 75°.
I think you flipped the sign of the "4ac" component in your quadratic formula. It should be 2 ± √3, not 2 ± √5. I did the same thing the first time through, really threw me off.
@@quigonkenny Sorry its sqrt(3) and not sqrt(5); tan(15°) = 2 - sqrt(3), and cotan(15°) = 2 + sqrt(3) It's a mistake when I copy my solution on the computer not a mistake when calculate the solutions.
To scale down from the large numbers. Let AC be length 4 instead of 92, a ration of 1/23. Area is proportional to side length squared. Therefore the area when AC = 4 will be 1058/(23 x 23) = 2 sq. units. Working from new smaller but exactly similar triangle. Let AB = length a, and BC = length b. Then ab/2 = 2, so ab =4, and b =4/a. Also, a^2 + b^2 = 4^2 Substituting for b = 4/a. a^2 + (4/a)^2 = 16. a^2 + 16/a^2 = 16. Let a^2 = u. u +16/u = 16. Multiply by u and rearrange. u^2 -16U +16 =0. Quadratic equation. u = 1.0718. But a^2 = 1.0718. a = 1.0353. Cos x = 1.0353/ 4 = 0.2588. x = 75 degrees.
Your approach also makes it easier to prove it with geometry/without trig. Solving u^2-16u+16=0, gives you the two sides for the triangle (after a fair mount of algebra). Then draw a 30-60-90 triangle DBC within triangle ABC. DB will be sqrt3 x the short side BC. One will then find that AB-DB=DC. Hence triangle ADC is isosceles. Then by exterior angle theorem 2x=30.
1/ Drop the height BH =h to AC. Label AH= m and HC=n, and assume that m>n WE have 1/2 h . AC= 1058 ----> h= 23 2/ By using the right triangle altitude theorem: sq h= m.n---> sq 23 = m.(92-m) ---> we have the equation: sqm -92m -529= 0----> m=85.84 so we have tan x= h/m= 23/(85.84)=0.2679 x= 15 degrees
Another way to calculate the value of m and tan x Once having the equation sq23= m.(92-m)----> sqm - 92m = - sq23 or (sqm - 2. 46m + sq46) -sq46 = - sq23-----> sq(m-46) = sq 46-sq23= (46+23)(46-23)= 69 x23= 23x23x3 -------> m-46= 23 sqrt3----> m=46+23 sqrt3= 23(2+sqrt3) tan x= h/m= 1/(2+sqrt3)
For the height, h, from B to the side AC: h = 2 * 1058 /92 The side AC is split by h into two sections, z and y, with y+z = 92 or y = 92-z, and y*z = h^2, which gives 92*z - z^2 = h^2 or z^2 - 92*z + h^2 = 0 From the two solutions of z, we can calculate the angle, x: arctan(x) = h/z When I've got time, I could add the numerical values.
Wow! Awesome! It's been a while since we've done Trig Identities and Angle Formulas. I absolutely love it! 🙂
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AB=b ; BC=a
1/2ab=1058
ab=2116
a^2+b^2=(92)^2=8464
(a+b)^2-2(2116)=8464
a+b=46√6; a+b=-46√6
b=46√6-a
a(46√6-a)=2116
a^2-46√6a+2116==0
a=23√6+23√2 a=23√6-23√2
b=46√6-23√6+23√2=23√6+23√2
b=23√6-23√2
cos(x)=92^2+(23√6+23√2)^2-(23√6-23√2)/2(92)(23√6+23√2)
cos(x)=(92)^2+(23√6-23√2)^2-(23√6+23√2)^2/2(92)(23√6-23√2)=75°
So : x=15°.; x=75° ❤❤❤ Thanks sir.
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Let the hypotenuse have length c and the sides a and b. We can simplify the algebra by factoring the hypotenuse length by 23, leaving c = 4 and the area by (23)², leaving (1/2)ab = 2 or ab = 4. The resulting triangle is similar, so has the same corresponding angles. Now, we take an educated guess that the triangle is either a special 30°-60°-90° or 45°-45°-90° right triangle. We try 30°-60°-90°. The sides are 1, √3, 2 which we multiply by 2 to make the hypotenuse c = 4, so 2, 2√3, 4. The product ab is (2)(2√3) = 4√3, which is not the required 4. We try 45°-45°-90°. The sides are √2/2, √2/2, 1 which scales up to 2√2, 2√2, 4. The product ab = (2√2)(2√2) = 8, which is not 4. So, we take another educated guess that the triangle is 15°-75°-90°, which has sides (√3 - 1), (√3 + 1), 2√2, and we scale up to (√2)(√3 - 1), (√2)(√3 + 1), 4. The product ab = (√2)(√3 - 1)(√2)(√3 + 1) = (2)(√3 - 1)(√3 + 1) = (2)(3 - 1) = 4. We are in luck! So, this is a 15°-75°-90° triangle and x = 15° or 75°.
Note that the 15°-75°-90° triangle is not designated as a "special" right triangle in geometry, while the 30°-60°-90° or 45°-45°-90° right triangles are. However, it appears frequently in problems, so it is worthwhile to know its side ratios.
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ab=2116, a^2+b^2=92^2=8464, (a+b)^2=8464+2×2116=(46sqrt(6))^2, a, b are given from (46sqrt(6)+-46sqrt(2))/2=23sqrt(2)(sqrt(3)+-1), thus b/a=(sqrt(3)-1)/(sqrt(3)+1)=(1-1/sqrt(3)))/(1+1/sqrt(3))=tan(45-30)=tan(15), therefore x=15 or 90-15=75.
1/2×(sqrt(3^2=1/2×(4-2sqrt(3))=2-sqrt(3), (1-1/sqrt(3))/(1+1/sqrt(3))
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.... Happy weekend to you, Let I AB I = S & I BC I = T ... S * T = 2 * 1058 = 2116 and S^2 + T^2 = 92^2 = 8464 ... [ 4 * 2116 = 8464 ] ... S^2 + T^2 = 4ST ... S^2 - 4ST + T^2 = 0 ... (S - 2T)^2 - 4T^2 + T^2 = 0 ... (S - 2T)^2 = 3T^2 ... S - 2T = +/- SQRT(3)*T ... S1,2 = (2 +/- SQRT(3))* T ... S1 = (2 + SQRT(3))*T1 v S2 = (2 - SQRT(3))*T2 ... recalling S*T = 2116 .... S1*T1 = (2 + SQRT(3))*(T1)^2 = 2116 .... (T1)^2 = 46^2/(2 + SQRT(3) ... T1 = 46*SQRT(2 - SQRT(3)) ... S1 = (2 + SQRT(3))*46*SQRT(2 - SQRT(3)) ... X1 = ARCTAN( T1/S1) = 15 deg. ... the same procedure for S2 & T2 ... I obtain X2 = 75 deg. ... S = { 15 deg. , 75 deg. } ... thank you sir for your continuing stream of interesting maths exercises and efforts, and the always clear accompanying explanation presentation .... best regards, Jan-W
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Let BC = a, AB = b and AC = c. Then we have ½ ab = 1058 and a² + b² = 92².
So a = 46√(2 − √3) and b = 46√(2 − √3) (2 + √3). Then a / b = 1 / (2 + √3).
So we can conclude that the angle x is tan⁻¹ (1 / (2 + √3)) = 15°.
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I propose something interesting (I hope so!):
Let'a name a = BC and b = AC. We have: a^2 + b^2 = 92^2 = 8464 (Pythagorean theorem in triangle ABC) and a.b = 2.area of triangle ABC = 2116.
Let's build a second degree equation whose solutions are a/b and b/a: This equation is (x -a/b).(x -b/a) = 0 or x^2 - (a/b + b/a).x + (a/b).(b/a) =0
So it is: x^2 - ((a^2 + b^2)/(a.b)).x+ 1 = 0 or x^2 - (8464/2116).x +1 = 0 or finally x^2 -4.x +1 = 0. Its solutions are 2 +sqrt(5) and 2 -sqrt(5)
Conclusion: tan(x) = a/b = 2 - sqrt(5) and x = 15° or tan(x) = a/b = 2 +sqrt(5) = cotan(15°)n = tan (90° - 15°) = tan(75°) and x = 75°.
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I think you flipped the sign of the "4ac" component in your quadratic formula. It should be 2 ± √3, not 2 ± √5. I did the same thing the first time through, really threw me off.
@@quigonkenny Sorry its sqrt(3) and not sqrt(5); tan(15°) = 2 - sqrt(3), and cotan(15°) = 2 + sqrt(3) It's a mistake when I copy my solution on the computer not a mistake when calculate the solutions.
Good Method 👍
Simple Method,
Better than Mine:
a"+b" = 92" & a•b = 2116 ?
(a")" - 92"(a") + 2116" = 0 😵
tan x = b/a = 2116/(a")
Two Solutions = 2 + √3 or 2 - √3 == 15° or 75° complex me 😜
Glad to hear that!
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A = bh/2
1058 = ba/2
b = (2/a)1058 = 2116/a ---- (1)
a² + b² = c²
b² = 92² - a²
b² = 8464 - a² ---- (2)
(2116/a)² = 8464 - a²
4477456/a² = 8464 - a²
4477456/y = 8464 - y
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Your method is the fastest, for no unnecessarily computing the length of adjacent sides.🎉🎉🎉
Excellent!
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Good method, well executed.
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X=15 degrees only, because AB = 92 cosx , cos 150 = -0.866, which means AB is negative,which is impossible.
To scale down from the large numbers.
Let AC be length 4 instead of 92, a ration of 1/23.
Area is proportional to side length squared.
Therefore the area when AC = 4 will be 1058/(23 x 23) = 2 sq. units.
Working from new smaller but exactly similar triangle.
Let AB = length a, and BC = length b.
Then ab/2 = 2, so ab =4, and b =4/a.
Also, a^2 + b^2 = 4^2
Substituting for b = 4/a.
a^2 + (4/a)^2 = 16.
a^2 + 16/a^2 = 16.
Let a^2 = u.
u +16/u = 16.
Multiply by u and rearrange.
u^2 -16U +16 =0.
Quadratic equation.
u = 1.0718.
But a^2 = 1.0718.
a = 1.0353.
Cos x = 1.0353/ 4 = 0.2588.
x = 75 degrees.
Thanks ❤️
Your approach also makes it easier to prove it with geometry/without trig. Solving u^2-16u+16=0, gives you the two sides for the triangle (after a fair mount of algebra). Then draw a 30-60-90 triangle DBC within triangle ABC. DB will be sqrt3 x the short side BC. One will then find that AB-DB=DC. Hence triangle ADC is isosceles. Then by exterior angle theorem 2x=30.
I was feeling whimsical today.
So,,, I am using horse feathers as the units. 😎
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Thank you! Appreciate how this is solved by identifying two equations. (And, all of the other videos too!)
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You are very welcome!
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1/ Drop the height BH =h to AC. Label AH= m and HC=n, and assume that m>n
WE have 1/2 h . AC= 1058 ----> h= 23
2/ By using the right triangle altitude theorem: sq h= m.n---> sq 23 = m.(92-m)
---> we have the equation: sqm -92m -529= 0----> m=85.84
so we have tan x= h/m= 23/(85.84)=0.2679
x= 15 degrees
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Another way to calculate the value of m and tan x
Once having the equation sq23= m.(92-m)----> sqm - 92m = - sq23
or (sqm - 2. 46m + sq46) -sq46 = - sq23-----> sq(m-46) = sq 46-sq23= (46+23)(46-23)= 69 x23= 23x23x3
-------> m-46= 23 sqrt3----> m=46+23 sqrt3= 23(2+sqrt3)
tan x= h/m= 1/(2+sqrt3)
Found that sinx = 1/2root(2-root3) but your way is much better
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Leta X = AB and Y = BC
1) X * Y = 1.058 * 2 ; X * Y = 2.116 ; X = 2.116 / Y or Y = 2.116 / X
2) X^2 + Y^2 = 8.464 (92^2)
3) X^2 + (2.116 / X)^2 = 9.216 ; X^2 + (2.116^2 / X^2) = 8.464 ; (X^4 + 2.116^2) = 8.464 * X^2 ; X^4 - 8.464X^2 + 2.116^2 = 0
4) Variable Change : Let X^2 = Y
5) Y^2 - 8.464Y + 2.116^2 = 0
6) Y = 566.98 or Y = 7.897
7) X = sqrt(Y)
8) X = sqrt(566.98) ~ 23,811
9) X = sqrt(7.897) ~ 88,865
Checking Solution :
Triangle Area = 1.058 =
Triangle Area = (sqrt(566.98) * sqrt(7.897)) / 2 ~ 1.057,998
X^2 + Y^2 = 92^2
566,98 + 7.897 = 8.464 : 8.463,98 ~ 8.464
x = arctan (23,811 / 88,865)
x ~ arctan(0,2679)
x ~ 15º
Answer:
Angle CAB (x) equal to 15º, considering that AB > BC.
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For the height, h, from B to the side AC:
h = 2 * 1058 /92
The side AC is split by h into two sections, z and y, with
y+z = 92 or y = 92-z, and
y*z = h^2, which gives
92*z - z^2 = h^2 or
z^2 - 92*z + h^2 = 0
From the two solutions of z, we can calculate the angle, x:
arctan(x) = h/z
When I've got time, I could add the numerical values.
Thanks ❤️
Nice 👍
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Thanks premath
You are very welcome!
Thanks dear ❤️
Whatever the solution is for x, 90-x is also a solution.
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Interesting sir!
Glad you think so!
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I loved this video! Thx
Glad you enjoyed it!
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Interestingly the two answers represent the two complementary angles in the triangle.
Thanks for the great video - i'm only getting the one angle for 2x of 30 degrees from sin-1(0.5) so how do i also get 150? thanks
Let's calculate x:
.
..
...
....
.....
Since the triangle ABC is a right triangle, we can conclude:
AB² + BC² = AC²
(1/2)*AB*BC = A(ABC)
(AB ± BC)²
= AB² ± 2*AB*BC + BC²
= AB² ± 4*(1/2)*AB*BC + BC²
= AB² ± 4*A(ABC) + BC²
= AC² ± 4*A(ABC)
(AB + BC)² = AC² + 4*A(ABC) = 92² + 4*1058 = 12696
⇒ AB + BC = √12696 = 46√6
(AB − BC)² = AC² − 4*A(ABC) = 92² − 4*1058 = 4232
⇒ |AB − BC| = √4232 = 46√2
First case: AB > BC
AB + BC = 46√6
AB − BC = 46√2
AB = (1/2)*[(AB + BC) + (AB − BC)] = (1/2)*(46√6 + 46√2) = 23√2(√3 + 1)
BC = (1/2)*[(AB + BC) − (AB − BC)] = (1/2)*(46√6 − 46√2) = 23√2(√3 − 1)
tan(x) = BC/AB = (√3 − 1)/(√3 + 1) = (√3 − 1)²/[(√3 + 1)(√3 − 1)] = (3 − 2√3 + 1)/(3 − 1) = 2 − √3
⇒ x = 15°
Second case: AB < BC
BC + AB = 46√6
BC − AB = 46√2
BC = (1/2)*[(BC + AB) + (BC − AB)] = (1/2)*(46√6 + 46√2) = 23√2(√3 + 1)
AB = (1/2)*[(BC + AB) − (BC − AB)] = (1/2)*(46√6 − 46√2) = 23√2(√3 − 1)
tan(x) = BC/AB = (√3 + 1)/(√3 − 1) = (√3 + 1)²/[(√3 − 1)(√3 + 1)] = (3 + 2√3 + 1)/(3 − 1) = 2 + √3
⇒ x = 75°
How to prove that tan(45°±30°)=2±√3:
tan(45° ± 30°)
= [tan(45°) ± tan(30°)]/[1 ∓ tan(45°)tan(30°)]
= (1 ± 1/√3)/(1 ∓ 1/√3)
= (√3 ± 1)/(√3 ∓ 1)
= (√3 ± 1)²/[(√3 ∓ 1)(√3 ± 1)]
= (3 ± 2√3 + 1)/(3 − 1)
= 2 ± √3
Best regards from Germany
Congratulations to the solution you have shown in the video. It is much easier.👍
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(92)^2=8464/1058°=√892.4 √89^1√2^1.√2^2 √1^√1 √1^√1 1^2 (x+1x-2)
√1058√ 5^√2;2^√29 ,√5^1^√2^1 2^√29^1 √1^√1 √1^√1 2^√1^1 2^1 (x+1x-2) √92. 2^√46 2^√23 2^√23^1 2^√1^1 2^1 (x+1x-2)
nice 👌
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