At a quick glance: Draw a perpendicular from C to form a right angled triangle ACD with AC = 2 and angle CAD = 180-120=60. then DA= 2 * cos(60) = 1 and DC= 2 * sin(60) = 1.73 = sqrt(3). DB^2 = 48- 3. DB=6.7 and AB=5.7.
Very quick: Use the "extended Pythagorean theorem" usually known by the formula a^2 = b^2 + c^2 - 2.b.c.cos(A) Here we have: BC^2 = AC^2 + AB^2 - 2. AB . AC . cos (120°) or: 48 = 4 + AB^2 - 4. AB. cos (90° + 30°) cos(120°) = -sin (30°) = -1/2, and let's note AB = x. Then we have: 48 = 4 + x^2 - 2x or: x^2 + 2.x - 44 = 0 The reduct delta is (-1)^2 + 44 = 45, so x = -1 - sqrt(45) or x = -1 + sqrt(45), the first solution beeing impossible as it is negative. Finally AB = x = -1 + sqrt(45) = -1 + 3.sqrt(5).
(3:05) You could have drastically simplified by directly solving for DB (trivially, in the head) being squared = 45, then subtracting 1 from the resulting root. Thus you would have circumvented the quadratic equation. BUT - as always: All roads lead to Rome!
Knowing sin(B), you can find cos(B)=√1-1/16=√15/4 , the angle is sharp. sin(C)=sin(180-120-B)=sin(60-B)=sin(60)cos(B)-cos(60)sin(B)=(3√5-1)/8. Applying the sine theorem, we find x=3√5-1. Now you can find the value with the required accuracy. When rounding the preliminary results, accuracy is lost
Or , in order to derive the law of cosines , by the use of a suitable construction we apply the Pythagorean theorem on a non right triangle . In other words , the law of cosines is an extension of the pythagorean to non right triangles .
Angle CAD is 60 degrees. therefore, CAD is a 30, 60, 90 triangle. So AD =1 and CD = sq root of 3 DB squared = CB squared minus CD squared = 48 - 3 = 45 DB = sq root 45 = 3 times sq root 5 approx = 6.708. AB = DB - 1 = 5.708
For the first method, the quadratic formula is unnecessary. From (x+1)^2+3=48 you get simply (x+1)^2=45, so x+1=sqrt(45)=sqrt(9*5)=3*sqrt(5), which yields x=3*sqrt(5)-1. For the second method, the Law of Cosines on triangle ABC gives you basically the same quadratic equation in x.
Comments that the problem can be solved by law of cosines are wrong. Check the equation c^2 = a^2 + b^2 - 2abCosC. The angle in equation must be the angle opposite to the unknown side, in other words the inclusive angle between the known sides. Obviously this is not the case in this problem. Interestingly, when 2 sides and 1 angle are given for equation of law of sines, the angle must not be an inclusive angle.(Check the equation to see this point.)This is basically a problem of congruent triangles. The triangle in the problem give SSA which may not be a positive congruent test. However, as the angle given here is an obtuse angle, this SSA is a congruence test as it gives only one solution. You can imagine that by rotating side CB (side opposite the given angle) about C. You can only get 1 intersecting point with side AB. (The other intersecting point on AB extension is outside the triangle hence is excluded.) Algebraic explanation is that there are 2 angles (acute angle in quadrant I, obtuse angle in quandrant II) for a positive value of sine derived from the equation for law of sines. The obtuse angle answer is excluded here as there can not be 2 obtuse angles in a triangle. When law of sines and law of cosines are used with underlying principles of congruence triangles, applicability and limitations become clear. Basically the equations may find the 4th parameter out of given 3 parameters of a triangle. (6 parameters of a triangle are 3 angles and 3 sides). With repetition, the 5th and 6th parameters are found hence a totally defined triangle, i.e. a congruent triangle proven by a congruence test with same set of 3 parameters in the equation.
Sorry, I was wrong in stating that law of cosines can not be used in this problem. Actually I want to emphasize laws of sines and cosines are closely related to congruence tests. SSA (as in this problem) is not always a congruence test as proven geometrically. This ambiguity is shown algebraically in equations of laws of cosines and sines. With law of sines, the equation may yield values of sine greater than 1 which is not allowed, or a valid sine value giving 2 possible angles in quadrant I and II (while an obtuse angle in quadrant II may be excluded like the case in this problem). The third possibility of 1 solution is the result of the non-inclusive angle being a right angle. With law of cosines, the equation is a quadratic equation that can give no real roots, 2 identical real roots or 2 positive real roots, 1 positive real root plus 1 negative real root (as in this problem), corresponding to the ambiguous results of law of sines. In conclusion, congruence tests SSS and SAS do not have a place in law of sines while SSA being an ambiguous congruence test can be dealt with by the laws of sines and cosines yielding no or multiple results in some conditions.
Another possible approach: Drop the height BH to AC (extended) We have: AHB is a 60-90-30 triangle. Just label AH= a so BH=a.sqrt3 and AB=2a We have sq(2+a) + sq (a.sqrt3)= sq(4sqrt3) Or sqa +a -11=O a=(3.sqrt5-1)/2 AB=3.sqrt5-1
In first method, when solving quadratic equation (x+1)² + (√3)² = (4√3)² you expand and move all terms to one side then use quadratic formula. However, another way to solve quadratics (which I sometimes find simpler) is by completing the square. Fortunately we are given the square already, so there's no need to complete it: (x+1)² + (√3)² = (4√3)² (x+1)² + 3 = 48 (x+1)² = 45 Since x+1 is a side length, we take positive square root: x + 1 = 3√5 x = 3√5 − 1
So if you wanted too trick us, after saying @ 0:35 if CB was less than 4×3^½ and less than the height of Triangle ABC then the Triangle ABC doesn't exist! In other words if A is acute and CB < Hieght or if A is obtuse and CB < or = Height and would have been verified after inspection using the Law of Sines and seen inequalities. Just sayin and havin fun. I am glad you included 2nd method. 🙂
I am going to use Trigonometry and Pythagorean Theorem. 1) Extend the Line AB to the left. 2) Draw a Line passing to point C and intersecting Line AB perpendicularly, call that point C'. 3) The angle [C'AC] is equal to 180º - 120º = 60º. 4) And the angle [ACC'] = 30º. 5) The angle [AC'C] = 90º 6) Knowing that cos(60º) = 1/2 = 0,5; we can conclude that AC' = 1, and CC' = sqrt(3) ~ 1,732 Now... 7) Let's call length AB, "x". 8) (1 + x)^2 + (sqrt(3))^2 = (4*sqrt(3))^2 9) 1 + 2x + x^2 + 3 = 48 10) Solving for x 11) x^2 + 2x - 44 = 0 12) x ~ - 7,7082 or x ~ 5,7082 13) Integer Solution is ~ 5,7082 My Answer: AB ~ 5,7082 linear units.
Hier noch eine Ergänzung zur gestrigen Aufgabe von PreMath. Bei der zweiten Methode über den Sinussatz wurde doch glatt mit gerundeten Zwischenergebnissen gerechnet, obwohl das gar nicht notwendig ist. Es geht nämlich auch exakt. Hier die Aufgabenstellung: Gegeben ist ein Dreieck mit den Seitenlängen a=4√3 und b=2 sowie dem Winkel alpha=120. Wie lang ist die Seite c? sin(alpha)/a = sin(beta)/b sin(120)/(4√3) = sin(beta)/2 (√3/2)/(4√3) = sin(beta)/2 sin(beta) = 1/4 sin(gamma) = sin(180-120-beta) = sin(60-beta) = sin(60)*cos(beta) - cos(60)*sin(beta) = sin(60)*√[1 - sin²(beta)] - cos(60)*sin(beta) = (√3/2)*√[1 - (1/4)²] - (1/2)*(1/4) = (√3/2)*√(1 - 1/16) - 1/8 = (√3/2)*√(15/16) - 1/8 = (√3/2)*√3*√5/4 - 1/8 = 3*√5/8 - 1/8 = (3*√5 - 1)/8 sin(beta)/b = sin(gamma)/c c = b*sin(gamma)/sin(beta) = 2*[(3*√5 - 1)/8]/(1/4) = 3*√5 - 1 Ich selbst hatte c mithilfe des Cosinussatzes berechnet, also mit folgendem Ansatz: a² = b² + c² - 2*b*c*cos(alpha)
Angle gamma= 180-120-beta=60-beta You could have continued with this identity: AB/sin(60-beta)=2/sin(beta)...but... I had been expecting you solve it with cosine law😮 the natural extension of Pythagorean theorem, right?
Well … your quadratic method is right on! However, you didn't really NEED to go through the quadratic expansion step. (𝒙 + 1)² = (4√3)² - (√3)² (𝒙 + 1)² = 48 - 3 𝒙 + 1 = √45 𝒙 + 1 = √(9 × 5) 𝒙 + 1 = 3√5 𝒙 = 3√5 - 1 𝒙 = 5.7082... Ta Da. No quadratic involved. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Can we say that i know it is wrong but can we say that cos 120 degrees = 2/baseand we can solve by using calculator i have request to tell me is it correct or wrong 😅
Si D es la proyección ortogonal de C sobre la alineación AB→ Ángulo DAC=60º→ DA=1 y DC=√3→ (4√3)² -(√3)²=DB²=45→ DB=√45→ AB=DB-DA =√45 -1 =5.7082 Gracias y un saludo.
Thank you anywayss😭 because I'm always on the first method team. I didn't think a question like that could be done without sincostan. Sines is killing me, and until now, i dont understand sines
I manged to do it in my head as ΔADC is 30°, 60° & 90° triangle. Gives AD=1 & CD=√3. Then BD²=BC² - CD² ⇒ 48 - 3 = 45. BD = √45 = ±3√5. Giving AB = 3√5 - 1. Done..Thanks.
We are masters of "careless mistakes", aren't we? In my first try - as i went on in the mood "Easy! Where is the problem?!" - i accidentally swapped the plus to a minus, fast forwarded to the end of the video and - by far not the first time - asked myself: Where the heck did i make the mistake? How is it even possible to make a mistake in a simple pythagorean expression!? Well: Carelessness is the dominant cause of all errors of mankind :D
No worries. No one is perfect! We are all lifelong learners. That's what makes our life exciting and meaningful! I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility! Thanks ❤️
Pin me im first to comment
You got it😀
Very good! We can use the Law o Cosines as well.
Thanks ❤️
At a quick glance: Draw a perpendicular from C to form a right angled triangle ACD with AC = 2 and angle CAD = 180-120=60. then DA= 2 * cos(60) = 1 and DC= 2 * sin(60) = 1.73 = sqrt(3). DB^2 = 48- 3. DB=6.7 and AB=5.7.
Thanks ❤️
Very quick: Use the "extended Pythagorean theorem" usually known by the formula a^2 = b^2 + c^2 - 2.b.c.cos(A)
Here we have: BC^2 = AC^2 + AB^2 - 2. AB . AC . cos (120°) or: 48 = 4 + AB^2 - 4. AB. cos (90° + 30°)
cos(120°) = -sin (30°) = -1/2, and let's note AB = x. Then we have: 48 = 4 + x^2 - 2x or: x^2 + 2.x - 44 = 0
The reduct delta is (-1)^2 + 44 = 45, so x = -1 - sqrt(45) or x = -1 + sqrt(45), the first solution beeing impossible as it is negative.
Finally AB = x = -1 + sqrt(45) = -1 + 3.sqrt(5).
Thanks ❤️
(3:05) You could have drastically simplified by directly solving for DB (trivially, in the head) being squared = 45, then subtracting 1 from the resulting root. Thus you would have circumvented the quadratic equation. BUT - as always: All roads lead to Rome!
Glad to hear that!
Thanks ❤️
Knowing sin(B), you can find cos(B)=√1-1/16=√15/4 , the angle is sharp. sin(C)=sin(180-120-B)=sin(60-B)=sin(60)cos(B)-cos(60)sin(B)=(3√5-1)/8. Applying the sine theorem, we find x=3√5-1. Now you can find the value with the required accuracy. When rounding the preliminary results, accuracy is lost
Thanks ❤️
The Pythagorean theorem is the law of cosines applied to a right angle.
Or , in order to derive the law of cosines , by the use of a suitable construction we apply the Pythagorean theorem on a non right triangle . In other words , the law of cosines is an extension of the pythagorean to non right triangles .
Thanks ❤️
Angle CAD is 60 degrees.
therefore, CAD is a 30, 60, 90 triangle.
So AD =1 and CD = sq root of 3
DB squared = CB squared minus CD squared = 48 - 3 = 45
DB = sq root 45 = 3 times sq root 5 approx = 6.708.
AB = DB - 1 = 5.708
For the first method, the quadratic formula is unnecessary. From (x+1)^2+3=48 you get simply (x+1)^2=45, so x+1=sqrt(45)=sqrt(9*5)=3*sqrt(5), which yields x=3*sqrt(5)-1. For the second method, the Law of Cosines on triangle ABC gives you basically the same quadratic equation in x.
Thanks ❤️
The quadratic formula is always unnecessary once one has understood how to complete the square.
Comments that the problem can be solved by law of cosines are wrong. Check the equation
c^2 = a^2 + b^2 - 2abCosC. The angle in equation must be the angle opposite to the unknown
side, in other words the inclusive angle between the known sides. Obviously this is not the case
in this problem. Interestingly, when 2 sides and 1 angle are given for equation of law of sines, the
angle must not be an inclusive angle.(Check the equation to see this point.)This is basically a problem of congruent triangles. The triangle in the problem give SSA which may not be a positive congruent test. However, as the angle given here is an obtuse angle, this SSA is a congruence test as it gives only one solution. You can imagine that by rotating side CB (side opposite the given angle) about C. You can only get 1 intersecting point with side AB. (The other intersecting point on AB extension is outside the triangle hence is excluded.) Algebraic explanation is that there are 2 angles (acute angle in quadrant I, obtuse angle in quandrant II) for a positive value of sine derived from the equation for law of sines. The obtuse angle answer is excluded here as there can not be 2 obtuse angles in a triangle. When law of sines and law of cosines are used
with underlying principles of congruence triangles, applicability and limitations become clear.
Basically the equations may find the 4th parameter out of given 3 parameters of a triangle. (6 parameters of a triangle are 3 angles and 3 sides). With repetition, the 5th and 6th parameters are found hence a totally defined triangle, i.e. a congruent triangle proven by a congruence test with same set of 3 parameters in the equation.
Thanks ❤️
Sorry, I was wrong in stating that law of cosines can not be used in this problem. Actually I want to emphasize laws of sines and cosines are closely related to congruence tests. SSA (as in this problem) is not always a congruence test as proven geometrically. This ambiguity is shown algebraically in equations of laws of cosines and sines. With law of sines, the equation may yield values of sine greater than 1 which is not allowed, or a valid sine value giving 2 possible angles in quadrant I and II (while an obtuse angle in quadrant II may be excluded like the case in this problem). The third possibility of 1 solution is the result of the non-inclusive angle being a right angle. With law of cosines, the equation is a quadratic equation that can give no real roots, 2 identical real roots or 2 positive real roots, 1 positive real root plus 1 negative real root (as in this problem), corresponding to the ambiguous results of law of sines. In conclusion, congruence tests SSS and SAS do not have a place in law of sines while SSA being an ambiguous congruence test can be dealt with by the laws of sines and cosines yielding no or multiple results in some conditions.
Another possible approach:
Drop the height BH to AC (extended)
We have: AHB is a 60-90-30 triangle. Just label AH= a so BH=a.sqrt3 and AB=2a
We have sq(2+a) + sq (a.sqrt3)= sq(4sqrt3)
Or sqa +a -11=O
a=(3.sqrt5-1)/2
AB=3.sqrt5-1
Thanks ❤️
In first method, when solving quadratic equation
(x+1)² + (√3)² = (4√3)²
you expand and move all terms to one side then use quadratic formula. However, another way to solve quadratics (which I sometimes find simpler) is by completing the square. Fortunately we are given the square already, so there's no need to complete it:
(x+1)² + (√3)² = (4√3)²
(x+1)² + 3 = 48
(x+1)² = 45
Since x+1 is a side length, we take positive square root:
x + 1 = 3√5
x = 3√5 − 1
Thank you!
You are very welcome!
Thanks ❤️
Good sharing sir❤❤❤❤
Many many thanks🌹
Respected Sir, I like your way of answering
Glad to hear that!
Thanks dear ❤️
So if you wanted too trick us, after saying @ 0:35 if CB was less than 4×3^½ and less than the height of Triangle ABC then the Triangle ABC doesn't exist! In other words if A is acute and CB < Hieght or if A is obtuse and CB < or = Height and would have been verified after inspection using the Law of Sines and seen inequalities. Just sayin and havin fun. I am glad you included 2nd method. 🙂
Excellent!
Thanks ❤️
I am going to use Trigonometry and Pythagorean Theorem.
1) Extend the Line AB to the left.
2) Draw a Line passing to point C and intersecting Line AB perpendicularly, call that point C'.
3) The angle [C'AC] is equal to 180º - 120º = 60º.
4) And the angle [ACC'] = 30º.
5) The angle [AC'C] = 90º
6) Knowing that cos(60º) = 1/2 = 0,5; we can conclude that AC' = 1, and CC' = sqrt(3) ~ 1,732
Now...
7) Let's call length AB, "x".
8) (1 + x)^2 + (sqrt(3))^2 = (4*sqrt(3))^2
9) 1 + 2x + x^2 + 3 = 48
10) Solving for x
11) x^2 + 2x - 44 = 0
12) x ~ - 7,7082 or x ~ 5,7082
13) Integer Solution is ~ 5,7082
My Answer:
AB ~ 5,7082 linear units.
Thanks ❤️
DB = sqrt(BC^2-CD^2)
= sqrt(48-3)
= sqrt(45)
= 3* sqrt(5)
AB+1 = 3*sqrt(5)
AB = 3*sqrt(5)-1
Thanks ❤️
It is easy to use the theorem of cosines.
Thanks ❤️
nice method of solving 🤨
Glad to hear that!
Thanks ❤️
Hier noch eine Ergänzung zur gestrigen Aufgabe von PreMath. Bei der zweiten Methode über den Sinussatz wurde doch glatt mit gerundeten Zwischenergebnissen gerechnet, obwohl das gar nicht notwendig ist. Es geht nämlich auch exakt. Hier die Aufgabenstellung: Gegeben ist ein Dreieck mit den Seitenlängen a=4√3 und b=2 sowie dem Winkel alpha=120. Wie lang ist die Seite c?
sin(alpha)/a = sin(beta)/b
sin(120)/(4√3) = sin(beta)/2
(√3/2)/(4√3) = sin(beta)/2
sin(beta) = 1/4
sin(gamma)
= sin(180-120-beta)
= sin(60-beta)
= sin(60)*cos(beta) - cos(60)*sin(beta)
= sin(60)*√[1 - sin²(beta)] - cos(60)*sin(beta)
= (√3/2)*√[1 - (1/4)²] - (1/2)*(1/4)
= (√3/2)*√(1 - 1/16) - 1/8
= (√3/2)*√(15/16) - 1/8
= (√3/2)*√3*√5/4 - 1/8
= 3*√5/8 - 1/8
= (3*√5 - 1)/8
sin(beta)/b = sin(gamma)/c
c = b*sin(gamma)/sin(beta) = 2*[(3*√5 - 1)/8]/(1/4) = 3*√5 - 1
Ich selbst hatte c mithilfe des Cosinussatzes berechnet, also mit folgendem Ansatz:
a² = b² + c² - 2*b*c*cos(alpha)
Großartig!
Danke ❤️
Angle gamma= 180-120-beta=60-beta
You could have continued with this identity:
AB/sin(60-beta)=2/sin(beta)...but...
I had been expecting you solve it with cosine law😮 the natural extension of Pythagorean theorem, right?
Thanks ❤️
Using Law of cosine -->
C² = A² + B² - 2ABcost
4²(3) = 4 + B² - 4B cos120
B² + 4B cos60 +4-48 =0
B² +2B -44 =0
B = (-2 + ✔️(4+4(44)) )/2
= -1 + 3✔️5 accepted
[ B= -1 - 3✔️5 rejected ]
.
You could have simply just used pythagoras to find DB=sqrt(16*3-3)=sqrt(45)=3*sqrt(5)
Then AB=DB-1=3*sqrt(5)-1
Thanks ❤️
Well … your quadratic method is right on! However, you didn't really NEED to go through the quadratic expansion step.
(𝒙 + 1)² = (4√3)² - (√3)²
(𝒙 + 1)² = 48 - 3
𝒙 + 1 = √45
𝒙 + 1 = √(9 × 5)
𝒙 + 1 = 3√5
𝒙 = 3√5 - 1
𝒙 = 5.7082...
Ta Da. No quadratic involved.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Thanks ❤️
If the angle opposite of side AB was given , it would be simple to apply the law of sines , and find length AB .
Thanks ❤️
(2)^24√3 =(4 DAC+64)=,,70°DAC (70DAC°+120°)=190DAC°(190DAC°-180°)=√10DAC° 5^2 (DAC+2DAC-5)
Can we say that i know it is wrong but can we say that cos 120 degrees = 2/baseand we can solve by using calculator i have request to tell me is it correct or wrong 😅
(3×root 5) - 1, may be
Thanks ❤️
Si D es la proyección ortogonal de C sobre la alineación AB→ Ángulo DAC=60º→ DA=1 y DC=√3→ (4√3)² -(√3)²=DB²=45→ DB=√45→ AB=DB-DA =√45 -1 =5.7082
Gracias y un saludo.
Thank you anywayss😭 because I'm always on the first method team. I didn't think a question like that could be done without sincostan. Sines is killing me, and until now, i dont understand sines
Thanks ❤️
😀 sines is killing me,too!
(4√3)^2=2^2+AB^2-2(2)(AB)cos(120)
AB=3√5-1=5.71 units.❤❤❤Thanks.
You are very welcome!
Thanks ❤️
I manged to do it in my head as ΔADC is 30°, 60° & 90° triangle.
Gives AD=1 & CD=√3. Then BD²=BC² - CD² ⇒ 48 - 3 = 45. BD = √45 = ±3√5. Giving AB = 3√5 - 1. Done..Thanks.
Excellent!
You are very welcome!
Thanks ❤️
Low of cosines also can solve
Thanks ❤️
the easiest way is to directly apply the law of cosines and you will get 5.71 square units as a result
Thanks ❤️
How are you applying the cosine rule ? Why the answer in square units ?
... Good day, I AB I = 8 * SIN(60 deg. - SIN^(- 1)(1/4)) = approx. 5.708 u. .... thank you for your presentation .... Jan-W p.s. SINE RULE (lol) ....
Thanks ❤️
asnwer=125 cm isit
3√5-1≈5,72
We are masters of "careless mistakes", aren't we? In my first try - as i went on in the mood "Easy! Where is the problem?!" - i accidentally swapped the plus to a minus, fast forwarded to the end of the video and - by far not the first time - asked myself: Where the heck did i make the mistake? How is it even possible to make a mistake in a simple pythagorean expression!? Well: Carelessness is the dominant cause of all errors of mankind :D
No worries. No one is perfect!
We are all lifelong learners. That's what makes our life exciting and meaningful!
I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility!
Thanks ❤️
Your computations can be shorter
Too difficult! CD = √3, AD = 1(Pythagor in ▲ACD), BD = √[(4√3)² - (√3)²] = 3√5 (Pythagor in ▲BCD).
AB = 3√5 - 1.
2/4 is not 1/8 but 1/2 (in the sines law)
Thanks ❤️
Es más fácil y comprensible el método 1, teorema de pitagoras. Saludos y gracias
You are very welcome!
Thanks ❤️
Sir please face reveal pls. pls. pls......
Cosine rule is better
Method 1 is easier
this is never math.
... X^2 + 2X - 44 = 0 is faster to solve by applying Completing the square ... (X + 1)^2 - 1 - 44 = 0 ... (X + 1)^2 = 45 ... X + 1 = +/- 3*SQRT(5) .... X1 = - 1 - 3*SQRT(5) < 0 (rejected) v X2 = I AB I = - 1 + 3*SQRT(5) is approx. 5.708 u. ...
Thanks ❤️
CalamariLobsterPotatoeWedgiesYellowBñueOysters....❤❤❤❤.