@ 0:46 when we were not going to use the Law of Cosines, I felt a tear drop from my face like I was at Opera Mathematica and felt that tragedy of losing a loved one. 🙂
I don't think it is necessary to calculate b. Once you know that a = 5/2 = half of the hypotenuse, you know that triangle ACD is a 30 60 90 triangle and that the angle DAC = 60°.
CD be a perpendicular to AB that touches AB at D Then CD = AC sin (x), DA = -AC cos ( x) CB^2 = CD^2 + BD^2 = AC^2 + AB^2 - 2 AB. AC cos ( x) Hereby cos (x) = - (7^2 - 3^2 - 5 ^2) /( 30) = - 15/30 = - 1/2 x = 120°
... Good day, 7^2 = 5^2 + 3^2 - 2*5*3*COS(X) ... COS(X) = - 1/2 ... [ QII, so obtuse angle X ] X = 120 deg. ... I know the application of the cosine rule is too easy to solve this exercise, but just to show the outcome is the same (lol) ... thank you for your continuing instructive maths efforts ... best regards, Jan-W
We can try the law of cosines: 7^2 = 5^2 + 3^2 - 2*5*3*cos(x) 49 = 34 - 30*cos(x) 15 = -30*cos(x) cos(x) = -0.5 We know cos(60) is 0.5, so we just want the mirror image angle of this: x = 120 degrees Q.E.D. Of course, this method only works without a calculator if you know your common angle trig function values.
My Oh my! You make it harder? Just Do it! Cosinus Law: 2•5•3•cos x = 5" + 3" - 7" = -15 cos x = -1/2 == -cos 60° = cos (180-60)° x = 120°! As simple as that^ 😗 29/03/24_7am_North Sumatra
1) The Sides of Triangle [ABC] are in a Sequence : 3 < 5 < 7, and 3 + 5 = 8 ; 3 + 7 = 10 ; 5 + 7 = 12 (8 < 10 < 12). And so the Angles are in a Crescent Sequence : ACB < ABC < BAC. 2) Calculating the Triangle Area by Herons Formula. Perimeter = 15 ; Semi Perimeter = 15/2 = 7,5 ; Area = 6,5 3) Being A' the point of perpendicularity over BC or height, height = AA' * 7 = 6,5 * 2 ; AA' = 13/7 ~ 1,86 4) 1,86 / 5 = sin(C) ; Thus, C = 21,84º 5) 1,86 / 3 = sin(B) ; Thus B = 38,32º 6) 180º - (21,84º + 38,2º) = 180 - 60,16º = 119,84º 7) Answer: I believe the right solution is x = 120º
Once we establish AD=5/2 (half the hypotenuse of triangle ADC), it immediately follows that ADC is 30-60-90 and angle x is 120 degrees. There is no need to find the length of DC (other than getting more exercise with the Pythagorean Theorem).
Generally, we will discover this method is equivalent to that applying cosine rule.5,3,7->a,b,c, so a^2-d^2=c^2-(b+d)^2=c^2-b^2-2bd-d^2, 2bd=c^2-a^2-b^2, d=(c^2-(a^2+b^2))/2b, cps y=(c^2-(a^2+b^2))/2ab, cos x=-cos y=((a^2+b^2)-c^2)/2ab. 🙃🙂🙃
You, too, can solve for the angle of a triangle without using trigonometry (as long as the triangle's dims conform to the dims of a "special triangle", which, of course, you have memorized)
Could this be an Orthogonal Projection problem which could use the theorem related to an orthogonal projection and obtuse angles. "In an obtuse-angled triangle, the square on the side subtending the obtuse angle is equal to the sum of the square on the sides containing the obtuse-angle together with twice the rectangle contained by one of those sides and the orthogonal projection of the other side upon it." BC²=AC²+AB²+ 2(AB)*(DA) 49=25+9+2(3)(DA). DA= 5/2 . . . DA equals half hypotenuse, therefore triangle CDA is 30-60-90, and angle DAC = 60°. Therefore angle CAB = 180° - 60°= 120°
Law of Cosines did the trick a= 3, b= 5, c= 7 c²= a²+b²-2ab cos C 49= 9+25-2(15)cos C 49= 34-30cosC 15= -30 cos C -1/2= cos C C= 120°, 240° (Note: Assume that the angle range is 0°≤C≤360°. However, since these are the triangle's interior angles, the only solution is 120°)
First, i went for the height from A to BC, which didn't deliver "nice" angles, but helped to get the idea of the task giver (used an arcsin calculation here). Then i looked for something giving the opposite angle of x and landet at the same as in the video. Nice exercise for cases where the quadratic terms vanish by themselves.
Today I will just present my "Modus Operandi" for this Problem. I will come later with a Resolution Proposal. 1) Calculate the Area knowing the Perimeter (= 3 + 5 + 7 = 15; Semiperimeter = 15/2), by Herons Formula. 2) Draw a Perpendicular Line from A to BC. Call it A'. 3) Divide the Area by 7. 4) Calculate the Height AA'. 5) Calculate the Angles ABC and ACB by means of Trigonometric Operations; Sine. 6) Find the angle Xº
it is a very good solution, I used an agebraic way of calculating sin of a part of an angle ane then used the formula of the sum of angles. I got the same result but the way you used is mor elegant.
By the Law of Cosines, 7² = 3² + 5² - 2(3)(5).cos(x). So 49 - 34 = - 30.cos(x). ∴ cos(x) = -1/2. Thus x = 120°. Also |AD|/|AC|= (5/2)/5 =1/2. So cos(180°-x)= 60°. Thus 180°-x=60°. ∴ x=120° again. Definitely not Olympiad level.
This method has a drawback - it cannot be applied as universal on a wide scale - change CB length from 7 to 6 units and our attempt ends in a fiasco... Almost no one can predict angle value starting to solve the problem. The Law of cosine is much more universal. But a calculator (trig tables) is required. 😅
OK, this time I just thought inside the "box", that means inside the triangle, so I could not find the necessary auxiliary lines. The method shown in the video is really nice.👍
Very elegant!
Thank you! Cheers!🌹
@ 0:46 when we were not going to use the Law of Cosines, I felt a tear drop from my face like I was at Opera Mathematica and felt that tragedy of losing a loved one. 🙂
Opera! wow
Thanks dear 🌹❤️
At 5:00, PreMath has found the value of a to be 5/2. So arccos(
Thanks ❤️
Yes, I noticed this too.
I don't think it is necessary to calculate b. Once you know that a = 5/2 = half of the hypotenuse, you know that triangle ACD is a 30 60 90 triangle and that the angle DAC = 60°.
Thanks ❤️
CD be a perpendicular to AB that touches AB at D
Then
CD = AC sin (x), DA = -AC cos ( x)
CB^2 = CD^2 + BD^2
= AC^2 + AB^2 - 2 AB. AC cos ( x)
Hereby
cos (x) = - (7^2 - 3^2 - 5 ^2) /( 30)
= - 15/30 = - 1/2
x = 120°
... Good day, 7^2 = 5^2 + 3^2 - 2*5*3*COS(X) ... COS(X) = - 1/2 ... [ QII, so obtuse angle X ] X = 120 deg. ... I know the application of the cosine rule is too easy to solve this exercise, but just to show the outcome is the same (lol) ... thank you for your continuing instructive maths efforts ... best regards, Jan-W
You are very welcome!
Thanks ❤️🌹
We can try the law of cosines:
7^2 = 5^2 + 3^2 - 2*5*3*cos(x)
49 = 34 - 30*cos(x)
15 = -30*cos(x)
cos(x) = -0.5
We know cos(60) is 0.5, so we just want the mirror image angle of this:
x = 120 degrees
Q.E.D.
Of course, this method only works without a calculator if you know your common angle trig function values.
My Oh my! You make it harder?
Just Do it! Cosinus Law:
2•5•3•cos x = 5" + 3" - 7" = -15
cos x = -1/2 == -cos 60° = cos (180-60)°
x = 120°! As simple as that^ 😗
29/03/24_7am_North Sumatra
Fantastic solution! Congrats!
Glad you like it!
Thanks ❤️
A very nice solution! Thank you PreMath! 🤩
You're welcome 😊
Thanks ❤️
1) The Sides of Triangle [ABC] are in a Sequence : 3 < 5 < 7, and 3 + 5 = 8 ; 3 + 7 = 10 ; 5 + 7 = 12 (8 < 10 < 12). And so the Angles are in a Crescent Sequence : ACB < ABC < BAC.
2) Calculating the Triangle Area by Herons Formula. Perimeter = 15 ; Semi Perimeter = 15/2 = 7,5 ; Area = 6,5
3) Being A' the point of perpendicularity over BC or height, height = AA' * 7 = 6,5 * 2 ; AA' = 13/7 ~ 1,86
4) 1,86 / 5 = sin(C) ; Thus, C = 21,84º
5) 1,86 / 3 = sin(B) ; Thus B = 38,32º
6) 180º - (21,84º + 38,2º) = 180 - 60,16º = 119,84º
7) Answer: I believe the right solution is x = 120º
Thanks ❤️
Once we establish AD=5/2 (half the hypotenuse of triangle ADC), it immediately follows that ADC is 30-60-90 and angle x is 120 degrees. There is no need to find the length of DC (other than getting more exercise with the Pythagorean Theorem).
Generally, we will discover this method is equivalent to that applying cosine rule.5,3,7->a,b,c, so a^2-d^2=c^2-(b+d)^2=c^2-b^2-2bd-d^2, 2bd=c^2-a^2-b^2, d=(c^2-(a^2+b^2))/2b, cps y=(c^2-(a^2+b^2))/2ab, cos x=-cos y=((a^2+b^2)-c^2)/2ab. 🙃🙂🙃
You, too, can solve for the angle of a triangle without using trigonometry (as long as the triangle's dims conform to the dims of a "special triangle", which, of course, you have memorized)
Thanks ❤️
Could this be an Orthogonal Projection problem which could use the theorem related to an orthogonal projection and obtuse angles.
"In an obtuse-angled triangle, the square on the side subtending the obtuse angle is equal to the sum of the square on the sides containing the obtuse-angle together with twice the rectangle contained by one of those sides and the orthogonal projection of the other side upon it."
BC²=AC²+AB²+ 2(AB)*(DA)
49=25+9+2(3)(DA).
DA= 5/2 . . . DA equals half hypotenuse, therefore triangle CDA is 30-60-90, and angle DAC = 60°. Therefore angle CAB = 180° - 60°= 120°
This was interesting. Law of Cosine would hit it out of the park. c=7.b=5 and a=3. x=120 degrees.
You are Genius sir
I'm just an ordinary...
Thanks dear ❤️🌹
Law of Cosines did the trick
a= 3, b= 5, c= 7
c²= a²+b²-2ab cos C
49= 9+25-2(15)cos C
49= 34-30cosC
15= -30 cos C
-1/2= cos C
C= 120°, 240°
(Note: Assume that the angle range is 0°≤C≤360°. However, since these are the triangle's interior angles, the only solution is 120°)
Thanks ❤️
Why would you assume range is 0° ≤ C ≤ 360°, when law of cosines is used to find the angle of a triangle (C = x), and this angle cannot exceed 180°?
@@MarieAnne. Oh yeah, Triangle's limits.
Then 120° is indeed the only solution.
Mine was for a full circle, thanks!
Muy bien profesor.
Excellent!
Glad to hear that!
Thanks ❤️
Thank you!
You are very welcome!
Thanks ❤️
First, i went for the height from A to BC, which didn't deliver "nice" angles, but helped to get the idea of the task giver (used an arcsin calculation here). Then i looked for something giving the opposite angle of x and landet at the same as in the video. Nice exercise for cases where the quadratic terms vanish by themselves.
Today I will just present my "Modus Operandi" for this Problem. I will come later with a Resolution Proposal.
1) Calculate the Area knowing the Perimeter (= 3 + 5 + 7 = 15; Semiperimeter = 15/2), by Herons Formula.
2) Draw a Perpendicular Line from A to BC. Call it A'.
3) Divide the Area by 7.
4) Calculate the Height AA'.
5) Calculate the Angles ABC and ACB by means of Trigonometric Operations; Sine.
6) Find the angle Xº
Thanks ❤️
Thank you
You are very welcome!
Thanks ❤️🌹
x=120°.
Thanks sir ❤❤❤
Great job!
You are very welcome!
Thanks ❤️
it is a very good solution, I used an agebraic way of calculating sin of a part of an angle ane then used the formula of the sum of angles. I got the same result but the way you used is mor elegant.
Excellent!
Thanks ❤️
x=120°
Without a calculator, the cos(x) works out very simply to -1/2 and we know that is cos(120).
By the Law of Cosines, 7² = 3² + 5² - 2(3)(5).cos(x).
So 49 - 34 = - 30.cos(x). ∴ cos(x) = -1/2. Thus x = 120°.
Also |AD|/|AC|= (5/2)/5 =1/2. So cos(180°-x)= 60°. Thus
180°-x=60°. ∴ x=120° again. Definitely not Olympiad level.
Thanks ❤️
25-x^2=49-(x+3)^2, 6x=49-9-25=15, x=5/2, thus angle=arccos(1/2)=60°, x=180-60°=120°.😅
Thanks ❤️
nice
thanks
My solution ▶
CB= 7
BA= 3
AC= 5
let AD= x
DC= y
If we apply the Pythagorean theorem to the triangle CAD:
AD²+DC²= AC²
x²+y²= 5²
If we apply the Pythagorean theorem to the triangle CBD:
BD²+DC²= CB²
(3+x)²+y²= 7²
9+6x+x²+y²= 49
x²+y²+6x= 40
x²+y²= 25 ⬆
⇒
25+6x= 40
6x= 15
x= 5/2
x²+y²= 25
25/4+y²= 25
y²= 25- (25/4)
y= √75/4
y= 5√3/2
let the angle A(CAD)= θ
tan(θ)= y/x
tan(θ)= (5√3/2) / (5/2)
tan(θ)= √3
θ= arctan(√3)
θ= 60°
x= 180°-60°
x= 120°
b) or, we may apply the cosine theorem to determine the angle x in this triangle CBA:
CB²= BA²+AC²-2*BA*AC*cos(x)
CB= 7
BA= 3
AC= 5
⇒
49= 3²+5²-2*3*5*cos(x)
49= 9+25-30cos(x)
15= -30cos(x)
cos(x)= -1/2
x= arccos(-1/2)
x= 120 °
Thanks ❤️
Magic! ✨
Excellent!
Thanks ❤️
Legal. A demonstração utilizada foi semelhante à própria demonstração da Lei dos Cossenos.
It was immediate with the law of cosines (cos(x) = -1/2). Morality: If you use only minimal tools, then be prepared to work a lot more. (But why not?)
Thanks ❤️
49=25+9-2*3*5cosx..15=-30cosx..cosx=-1/2..x=120
Thanks ❤️
This method has a drawback - it cannot be applied as universal on a wide scale - change CB length from 7 to 6 units and our attempt ends in a fiasco...
Almost no one can predict angle value starting to solve the problem.
The Law of cosine is much more universal. But a calculator (trig tables) is required. 😅
Thanks ❤️
Why is a calculator required?
@@murdock5537 to caclulate arccos
arccos((25+9-49)/(2×5×3))=arccos((-15)/(30))=arccos(-1/2)=120. This is the way of solution, not expected😂.
Thanks ❤️
We obtain the same answer using the cosine rule, with exactly the same equations.
Thanks ❤️
cb^2 = ab^2+bc^2-2*ab*ac*cosX
=
Thanks ❤️
It is a typical exercise for application ofcosine formula😮, you mean we have to solve it without any trigonometric tool?😅
I'm trying to share multiple tools...
Thanks ❤️
We can also do this as simple by law of cosine sir..btw this method u showed is a beautiful one..
yes, calculators are not required even using cosine rule.
But use of Pythagorean theorem is desired @@anthonycheng1765
Yes, definitely
Thanks ❤️
I guessed it on first sight. There are only so many angles you can find without trigonometry....
Thanks ❤️
@@someonespadre ok, i will amend my sentence to "There are only so many angles you can expect people to have lodged in their memory"
Arcsin(11/14)+arcsin(13/14)
Thanks ❤️
127
По теореме косинусов быстрее получится.
Истинный
Thanks ❤️
Х=120 градусов.
Just used cosine rule
Thanks ❤️
Let's find x:
.
..
...
....
.....
Why not using the law of cosines? Let's have a try:
BC² = AB² + AC² − 2*AB*AC*cos(∠BAC)
7² = 3² + 5² − 2*3*5*cos(x)
49 = 9 + 25 − 30*cos(x)
15 = −30*cos(x)
−1/2 = cos(x)
⇒ x = 120°
Best regards from Germany
OK, this time I just thought inside the "box", that means inside the triangle, so I could not find the necessary auxiliary lines. The method shown in the video is really nice.👍
Thanks dear 🌹❤️
định lý cosin...
Thanks ❤️
Exceptional
Glad to hear that!
Thanks ❤️
Oral
Got answer in 20 sec
Thanks ❤️
(3)^2 =9 (5)^2=25 (7)^2=49 {9+25+49=}83 3(15°)=45° 3(15°) =45° {45°+45°+90°} =180° {83-180°}=√97° 97^1 (ABCx-1ABCx-97)
Thanks ❤️
φ = 30° → cos(2φ) = sin(φ) = 1/2 → cos(4φ) = -cos(6φ - 4φ) = -cos(2φ) = -1/2
49 = 34 - 30cos(x) → cos(x) = -1/2 → x = 4φ
or: φ = 30°; ∆ ABC → AC = 5; AB = 3; BC = 7 = BQ + CQ = (7 - a) + a → sin(AQB) = 1 → AQ = h
BAC = x = QAC + BAQ = γ + δ → ACQ = α → QBA = β →
sin(α) = cos(γ) → cos(α) = sin(γ) → sin(β) = cos(δ) → cos(β) = sin(δ)
h = √(9 - (7 - a)^2 ) = 25 - a^2 → a = 65/14 → 7 - a = 33/14 →
sin(α) = cos(γ) = 3√3/14 → cos(α) = sin(γ) = 13/14; sin(δ) = cos(β) = 11/14 → cos(δ) = sin(β) = 5√3/14 →
sin(x) = sin(γ + δ) = sin(γ)cos(δ) - sin(δ)cos(γ) = (13/14)(5√3/14) + (11/14)(3√3/14) = √3/2 →
sin(x) = 2φ or (6φ - 2φ) = 4x → x > 3φ → x = 4φ
or: φ = 30°; h = √(9 - (7 - a)^2) = 25 - a^2 → a = 65/14 → 7 - a = 33/14 → h = 15√3/14 →
area ∆ ABC = (1/2)7h = 15√3/4 = (1/2)sin(x)(3)(5) → sin(x) = √3/2 → x = 4φ
Thanks ❤️
Sir this method works only for special angles. So I feel not worth your effort. I love your channel otherwise!
Thanks ❤️