Find the angle X | How to Solve this Tricky Geometry problem Quickly
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- čas přidán 25. 04. 2022
- Learn how to find the angle X in the given triangle. Solve this tricky geometry problem by using isosceles triangle property, straight angle, and 30-60-90 special triangle rule.
Today I will teach you basic tips and hacks to solve this tricky geometry problem in a simple and easy way. Step-by-step tutorial by PreMath.com
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Find the angle X | How to Solve this Tricky Geometry problem Quickly
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Learn how to find the angle X
How to Solve this Tricky Geometry problem Quickly
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Excellent method. Very interesting problem.
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome, John 😀
@@PreMath clever.
Hi
Nice solution.I also made a different solution but I can't post photos here
@@ufukkoyuncu3408 Great minds abound. Thank U 4 thinking.
czcams.com/video/HPAvvn87szM/video.html
Very clever! I solved it by trigonometry, getting tan(x) = 1/(2-sqrt(3)). The video method is better though
So nice of you.
Thank you for your feedback! Cheers!
You are awesome. Stay blessed 😀
How u got tan(x) ???
Sir-ji, it is not obvious that triangle ADP will be a right-angled triangle. We must use the Cosine Rule to deduce this.
|AP|² = |AD|² + |DP|² - 2.|AD|.|DP|.cos (D) = (2² + 1²) - 2.(2).(1).cos(60) = 5 - 4.(1/2) = 3. Thus |AP| = √3. Since
(2)² = 3 + 1 = (√3)² + (1)², |AD|² = |AP|² + |DP|². But |AD|² = |AP|² + |DP|² - 2.|AP|.|DP|.cos (P) by the Cosine Rule.
So 2.|AP|.|DP|.cos (P) = 0. Hence cos(P) = 0 and so P must be a right angle. The rest will now be okay. Jai Hind!
Any triangle with a 60° angle with adjacent sides with a 2/1 ratio is a 30/60/90 triangle, so it is in fact apparent and you don't need all that extra work to prove this. This is explained in the video.
@@MutatedWaffle77but isn’t the proof of that fact u cite due to the law of cosines or some equivalent statement?
Thanks, I agree. This is geometry, a science not faith or religion, you have to be able to prove every step. Here you proved it using cosine law.
Imagine you have a 2 unit line, at one end you can draw a circle with center at one end and radius on 1 unit. All the points on the circle could make a triangle, that has one side on unit and other other side 2 unit. Now if you put another specification that the angle between these two lines to be 60 degrees than you end up with two triangle, which can be proven to be right triangle, as shown by Ramu above. Here is statement of problem:
Prove that any triangle with sides 1 and 2 unit in length, and at 60 angle with respect to each other makes a right triangle.
You can prove that using cosine law as shown by Ramu or other methods.
@@MutatedWaffle77
Any right-triangle with a 2/1 longest-side to third-longest-side ratio, will have 60° as the included angle.
But the converse of this statement is not obvious. So I did need all that extra work to prove it. Just ask
the Premath professor himself - and I am pretty sure, he will tell you so. Pax vobiscum.
@@MutatedWaffle77 Yup, it's called a special triangle. It's basic geometry knowledge, but it's not always taught (which I find very lame).
Perfect angle chasing. Your problems and solutions are getting better day by day. Keep going! ❤️🇧🇩
That's really a very nice solution - thanks a lot!
Very nice problem for both algebraic (I did it first) and geometric solutions. I used a similar procedure, created equilateral triangle (side=1) on 60° angle to get point P. Due to the right-left symmetry, AP = BP applies. After completing all angles and constructing the second isosceles triangle BPC, I get AP = BP = CP , so P is the center of the circumscribed circle. We know the value of the angle BPC = 150° (the central angle), so the value X of the sought (circumferential) angle is half : X =
Nice question, professor. Bring us more of geometry problems, your resolution method is the best to learn this beautiful math area too. Thanks from Brazil 🇧🇷
More to come! Keep watching.
So nice of you.
Thank you for your nice feedback! Cheers!
You are awesome 😀
Love and prayers from the USA!
czcams.com/video/HPAvvn87szM/video.html
Thank you for the great problem!
素晴らしい！
見事な解決法です😃
それを聞いてうれしいです！
ご意見ありがとうございます！乾杯！
あなたは素晴らしいです😀
アメリカからの愛と祈り！
The rigid problem has been nicely solved. Superb!
Excellent execution 🙏❤️ Dear sir as always superb creation 🙏🙏🙏❤️❤️
Very impressive. Leveraging on the properties of the 30-60-90 triangle by selecting point P was very clever. I couldn‘t figure it out!
So nice of you, Philip.
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome 😀
Here is how to demonstrate (circle geometry) that APD is right angle triangle.
Let O middle of AD ---> OA = OD = 1 ----> ODP is isosceles (OD=DP=1) and angle ODP = 60 ----> triangle ODP is equilateral --->
OD=OP=PD = 1 ---> the circle with center O and diameter AD is circumscribed to triangle ADP ----> angle APD is right at P
Good morning Master
A Hug From Brazil
Congratulations
Thank you so much for the classes
Wow 😲 thanks a lot sir. It was very to understand 💖
Please keep bringing more geometry problems
I am little bit weak in it
Very elegant solution, thank you for sharing.
Using trig, and 120 degree, 45 degree and length 1, and the law of sine to get 3.346 for side facing 120 degree.
Then use the law of cosine with values , 3 (2+1), 45 degree and 3.346 to 75 degree for x. answer 75 degree
Regarding the questions about the 90 degree angle formed, can't one also justify the conclusion because the triangle is similar to a 30-60-90 triangle with side lengths 1, 2, sqrt(3) by SAS similarity where 60 is the included angle.
The crux was to look for an isosceles triangle which would be helpful. Nice!
The second bit was to spot the special 2/1 triangle with a fitting 60 angle, so you can prove the right angle. From there, the rest follows easily.
I did it using the CH height in triangle ADC.
CHB is isosceles in H given its 45 degrees angles (easy to find via angle sums).
Then you calculate angle tangents using lengths calculations.
You get tan(x) = 2+sqrt(3).
Knowing tan(60) = sqrt(3), you guess x is bigger, but related to 60 or 30 degrees (the root(3)), so you try 60 + 15.
Using the tan sum formula: tan(60+15) = tan(90-15) = 1/t15
You get t15^2 -2root(3)t15 - 1 = 0, so t15=2-root(3) = 1/(2+root(3)) = 1/tx
=> x=90-15 = 75
Let |CD| = b units, angle ACD = y & angle BCD = z. Then x + y =120° = x, so x = 120° - y & z = 15°. Now by the Sine Rule, sin(15°)/1 = sin(45°)/b and sin(y)/2 = sin(x)/b. So from the 1st equation, 1/b = sin(15°)/sin(45°). Substituting in the 2nd eq., we get sin (y)/2 = sin(120° - y).sin(15°)/sin(45°). So sin(45°).sin(y) = 2.sin(15°).sin(120°) = 2.sin(15°).sin(60+y), since the sines of supplementary angles are equal. Expanding we get,
sin(45°).sin(y) = 2.sin(15°).[sin(y).cos(60°) + cos(x).sin(60°)]. So [sin(45°) - 2.sin(15°).cos(60°)].sin(y) = [2.sin(15°).sin(60°)].cos(y). Thus sin(y)/cos(y) = 2.sin(15°).sin(60°) / [sin(45°) - 2.sin(15°).(1/2))] = 2.sin(15°).cos(30°) / [sin (45°) - sin(15°)]. But sin(45°) - sin(15°) = 2.sin {(45°-15°)/2}.cos{(45°+15°)/2}. So tan(y) = 2.sin(15°).cos(30°) / 2.sin(15°).cos(30°) = 1. Hence y = 45° and x = 120°- y = 75°.
----------------------
EXTRA: Also b = sin(45°)/sin(15°) = sin(15°) = [(√2)/2] / [(√6 - √2)/4] = [(√2)/2].[(√6 + √2)].(4) / [(√6 + √2).(√6 - √2)] = [(√12 + √2.√2)].(4) / [(2).(6 - 2)] = 4.(2).(√3 + 1)/8 = 1 + √3. The other sides by can be found by using the Sine Rule. |AC| = sin(45°).|AB|/sin(60°) = [(√2/2]. 3 / (√3 / 2) = √6 and |BC| = sin(75°).|AB|/sin(60°) = 3.[(√6 + √2)/4] / [(√3 / 2] = (3√2 + √6) / 2. Perhaps I should stop saying, Trig. is king. Trig. is actually Queen!
.
thank you for this easy approach.
What a great explanation!👏👏👏👏👍
So nice of you.
Thank you for your feedback! Cheers!
You are awesome, Mahalakshmi 😀
Love and prayers from the USA!
I solved this question in a different way by using trigonometry..
draw CF perpendicular to AB.
in triangle CED
TAN 60 = CE/ED
CE= √3ED -----(1)
In triangle CEB ,
TAN 45 = CE/ BE
1 = CE /( ED +BD)
1 = √3 ED / (ED + 1)..... [USING 1]
ED+1 = √3ED
ED(√3-1) = 1
ED = 1/(√3-1)
CE = √3/(√3-1)
AE + ED = AD
AE + 1/(√3-1) = 2
AE = √3( 2 - √3)/ (√3-1)
In triangle CEA ,
TAN X = CE/AE
TAN X = √3/(√3-1) ÷ √3(2-√3)/(√3-1)
= 1/( 2-√3)
= 2+√3
hence x = 75 degrees
Thanks 👍
It is very much Praiseworthy,and I also waited for ur video since I get to know about this channel.
So nice of you.
Thank you for your feedback! Cheers!
You are awesome, Saurav 😀
Love and prayers from the USA!
@@PreMath is it possible for to create the same moment with u???
Thanks for video. Good luck sir!!!!!
So nice of you.
Thank you for your feedback! Cheers!
You are awesome 😀
really good problem!
Another great solution👍
Thank you so much sir😊😊
You are very welcome.
So nice of you.
Thank you for your feedback! Cheers!
You are awesome 😀
I was actually able to do this in my head.
Very well thought out, I think this is good for understanding other methods for solving, but personally I don't think this would be a good approach on a test (if it is timed). It could be lowered to less steps with cosine I believe. Good quality video nevertheless 👍
Drop a perpendicular line from C to intersect AB at E.
Let AE=t and EB=3-t
Using the sine rule in triangle CDB,
with angle CBD=45 deg, CE=EB=3-t,DB=1
CD=sqrt2*EB,since triangle CEB is (90,45,45)
CD/sin45=CB/sin120
CD=2(3-t)/sqrt3 after working out.
Now consider triangle CED,
ED=2-t,CE=3-t ,CD=2(3-t)/sqrt3
Use Pythagoras,
CD^2=CE^2+ED^2
(2(3-t)/sqrt3)^2=(3-t)^2+(2-t)^2
2t^2-6t+3=0
t works out at t=(3+/-sqrt3)/2
since t=(3+sqrt3)/2>2
t=(3-sqrt3)/2 is the value of t to be used,
Now, consider triangle CAE,
tanx=CE/AE=(3-t)/t =(3-(3-sqt3)/2)/((3-sqrt3)/2)=(3+sqrt3)/(3-sqrt3)
=(3+sqrt3)(3+sqrt3)/(3-sqrt3)(3+sqrt3)
=(12+6sqrt3)/6=(2+sqrt3)
tanx=(2+sqrt3)
x=arctan(2+sqrt3)=75 degrees. This is our answer.
Sir I solved this by trigonometry
We have the angle on the right dide of 60 deg angle because of straight angle=is 120 deg
then the third angle of the right side triangle=15 deg
bc 15+120+45=180
now we have all angles and 1 side which is=1
calculate the length of the midline by sin rules
Then we draw a perpendicular from the upper angle on straight horizpntal line. No we have another 30 60 90 triangle thus we can calculate easyily the base and the perpendicular lengths then we can subtract the base from 2 thus we can find the base of the last left triangle where we already have the perpendiculare of the triangle now we can apply the cose rule to calculate the angle x
From the external angle theorem for traingles, the /_ ACD = 60 deg = /_ BCD + /_ CBD
= /_ BCD + 45; from this equation, /_ BC = 60-45 = 15 deg. Consider the triangle CBD, and et l represent the lenght of the line CD. By aplying the law of sines, we get:
BD/sin 15 = CD/sin 45
Substitute BD = 1 and CD = l
1/sin15 = l/sin45 => l = sin 45/sin 15 = (1/sqrt 2)/((sqrt 3 - 1)/(2*sqrt 2)). By rationalizing the denominator and simplifying this equation, we get:
l = 1 + sqrt 3 -------------- (1)
Extend the line AB toe the left to a point P, such that DP = CD . Thus, in the triangle CDP,
DP = CD = l = 1 + sqrt 3, therefore the angle DPC = 60 deg. Since two of the angles in triangle CDP are equal to 60, the third (/_ PCD) is also 60 deg, making this an equilateral traingle. Thus, the line PC = DP = CD = l = 1 + sqrt 3 ....... (2)
Also, the line segment AP = DP - AD = 1 + sqrt 3 - 2 = sqrt 3 - 1 ---------- (3)
Consider the triangle CAP, and use the cosine law:
AC^2 = AP^2 + PC^2 - 2 * AP * PC * cos /_CAP
= (sqrt 3 - 1 )^2 + (sqrt 3 + 1)^2 - 2 * (sqrt 3 - 1) * (sqrt 3 + 1) * cos 60 deg
= 2(3^2 + 1^2) - 2 * (3^2 - 1^2) * (1/2) = 2(4) - 2 = 8 - 2 = 6
Thus, AC = sqrt 6.
Consier the triangle ACD, and use the law of sines:
CD/sin x = AC/sin 60
(sqrt 3 + 1)/sinx = sqrt 6/(sqrt 3/2)
This equation can be manipulated and simplified to give:
sin x = (sqrt 3 + 1)/(2*sqrt 2) = cos 15 deg = sin (90-15) = sin 75 deg.
The the /_ x = 75 deg.
I solved it by constructing a 60-30-90 triangle DEC with E along side AD and then calculating the tangent of x, but this solution is so much nicer and more elegant. Cheers!
Super job!
Thank you for your feedback! Cheers!
You are awesome, Danny 😀
I did the same method, tangent x and then arctangent (knowing CE and AE)
Even though it may be true. You never offered any proof that
This is the comment i am looking for. 😁
Yep, he kinda skipped over how he got the right angled triangle APD.
czcams.com/video/HPAvvn87szM/video.html
@@davidwood6283 He should have done better...
straight line AB=180
forming a triangle with the straight line with the given angles
60+45=105
subtracting the angles formed by the straight line=105 from 180
180 - 105 = 75
(😅 I just hope this is acceptable)
Let A be the origin, AB be the x-axis, B = (x₁,y₁), D= (x₂,y₂), and C = (x₃,y₃). By the Point-slope formula, we have y = m.(x-x₁). So the "-45°" degree line has equation y = tan(-45°).(x-3) = 3-x. For the "-60°" degree line, we have y = m₂.(x-x₂). So y = tan(-60°).(x-2)= (-√3)(x-2) = (2-x).√3. These two lines intersect at the point C=( x₃,y₃) when 3-x = (2-x)√3. So x(√3 -1) = 2√3 - 3. Hence x₃ = (2√3 - 3)/(√3 -1) = (2√3 - 3).(√3 +1)/[(√3 -1).(√3 +1)] = (3-√3)/2. So y₃ = m₁(x-x₃) = (-1).(3-√3)/2 = (3+√3)/2.
Hence tan(∠CAD) = y₃/x₃ = (3+√3)/(3-√3) = (12+6√3)/6 = 2+√3. So ∠CAD = tan⁻¹(2+√3) = 75°. Why? Well, tan(75°) = tan(45°+30°) = [tan(45°)+tan(30°)] / [1- tan(45°).tan(30°)] = [1 + 1/√3]/[1 - 1/√3] = [√3 +1]/[√3 -1] = [(√3 +1).((√3 +1)] / [(√3 -1).(√3 +1)] = [4+2√3]/2 = 2+√3. The calculations were a little bit ugly - but the idea is straight-forward and required no special trick or construction.
.
Great! At the beginning. You refer the AB is divided in the ratio of 1:2. Later, you say the length as is 1 unit. Thats where i am not clear. Anyway, this is a roundabout method with assumption.
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
It is clear from many comments here that APD = 90 is not well known universal constant. I think you could explain this in your videos.
I don't understand where this conclusion came from either ...
If Point P exists, and AP is perpendicular to CD, than angle PAD = 30 degree, angle DPA = 90 degree. Therefore, AP = √3 (due to 30-60-90-triangle). That means that DP must be 1 = BD. He should've explained it...🙂
my method: drop a perpendicular from C to AB and name the point X. now we know that DB = 1 , let XD=a and CX=b hence tan 45 = b/(a+1) and tan 60 = b / a , on solving we get a = (root3+1)/2 and b = root3/2(root3+1) , now we know that AD = 2 hence AX = AD-XD = root3/2(root3-1), now we previously calculated CX = root3/2(root3+1) hence tanx = CX/AX = (root3+1)/(root3-1) hence giving x =75 degrees
Bravo again!
So nice of you.
Thank you for your feedback! Cheers!
You are awesome, Nenad 😀
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my 10 cent solution:
apply sine law on triangle CBD: sin 15/1 = sin 45/CD = sin120/BC. Using a slide rule with trig function, we get CD=2.73 and BC=3.34.
then on triangle ACD, apply the cosine law: AC^2=AB^2+BC^2 -2xABxBCxcos45, where AC=2.45. AB is given, 2+1= 3.
lastly, on triangle ACD, back to slide rule to apply sine law: sin X/2.73 = sin 60/2.45. X= 75 degrees
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal '' '' '' '
A variation on MarieAnne's alternative solution, using only one new variable. Drop a perpendicular from point c to AB and label the point of intersection as point E. ΔADE is a special 30°-60°-90° right triangle. Designate the length DE as x. Then, length CE = (√3)(x) by way of ratio of sides of the special 30°-60°-90° right triangle. Note that ΔABE is a special 45°-45°-90° right triangle, so sides CE and BE have equal length. Length BE = x + 1 and length CE = (√3)(x), so x + 1= (√3)(x). Solving, x = 1/(√3 - 1). Length CE = (√3)/(√3 - 1) = (√3)(√3 + 1)/((√3 - 1)(√3 + 1)) = (3 + √3)/ 2. Length AE = 2 - Length DE = 2 - x = 2 - 1/(√3 - 1) = 2 - (√3 + 1)/((√3 + 1)(√3 - 1)) = 2 - (√3 + 1)/2 = (4 - (√3 + 1))/2 = (3 - √3)/2. So, in ΔACE, the ratio of side opposite
Weldone son !
I like this level question and like to have faster solution !!
Professor: your method is fantastic.
I got the x=75 using m-n theorem
Excellent!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome, Sandanadurai. Keep rocking 😀
I used the law of sines and the law of cosines and a calculator. Angle DBC is given with 45. Angle BDC is 180 - 60 = 120. Angle BCD is 180 - 120 - 45 = 15. 1 / sin 15 = CD / sin 45 -----> CD is approximately 2.732051. AD is given as 2. (AC) ^2 = (AD) ^2 + (CD)^2 - 2 * (AD) * (CD) * cos 60. I came up with AC = sq-rt (6). Sq-rt (6) / sin 60 = 2.732051 / sin X. Sin X is approximately 0.965926. X = 75 degrees.
got it, thanks for the challenge bro
You are very welcome.
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome 😀
3:14 sir here we can also say that ADP is a right triangle because obtuse angle PDB =120° SO angle DAP =120-60=30° hence DPA =90°
The sum of the opposite interior angles is 120, not 60 + x; although in this particular case it happens to be true.
What does it mean 2:1?
that's nice
Another way to do it.
Draw a Circumcircle and extend CD to meet the circle at F. Raise a perpendicular bisector from AB at E to meet CD at P.
Since ED = ½
PD = 1 = DB
Hence, ∠DBP = ∠DPB = 30°
Chord FB subtends 15° at C and 30° at P, implying P is the center of the circle.
Since ∠CPB = 150°
Chord CB subtends 75° at A.
Image here for reference
czcams.com/video/pHyuSWoVLfw/video.html
awesome!
Impressive use of cords angles in a circle and its reciprocal. Most elegant proof. Kudos.
Question: why You asume P is 90 degrees? Is that a rule, When You have 60 degrees and proportions 2 - 1???
How do I learn construction or engineering work in difficult questions to reach a solution, please🙏👍❤️
the most difficult thing here is the additional construction
Nice
Thnku
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome, Pranav 😀
Nice a task! For a while all your geometric tasks are only in planimetry. Could you show any task in stereometry in your lessons? Thank you in advance, sir! God bless and save you!
Great suggestion!
So nice of you.
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You are the best, Anatoliy 😀
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@@PreMath u live in US?
@@TOMGEMANAR
Yes!
@@PreMath are u indian
V nice explanation. Construction part v imp
Glad to hear that!
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You are awesome, Niru 😀
awesome,
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I used the sine rule to find BC=3.346 units then the cosine rule to find AC=2.449 units then finally the cosine rule to find X=75°
The author is quite adept at triangles involving 60,30 and 45 degrees (very impressive). Unfortunately, the world of triangles is not confined to these angles. Good old fashioned trigonometric equations are a sure fire way to solve this problem for angles not 60 and 45 degrees. I did it in six lines of short equations.
Nice video thanks
So nice of you.
Thank you for your feedback! Cheers!
You are awesome. Keep it up. 👍
Love and prayers from the USA! 😀
Sir, you jumped over deriving the right angle. One has to describe such a derivation in more detail, otherwise the procedure does not match your HABIT of explaining many simpler derivations as easy.
1. Calculate length of BC from sine law in triangle CDB
|BC|/sin(120)=1/sin(15)
|BC| = sin(120)/sin(15)
|BC| = sqrt(3)/2*4/(sqrt(6)-sqrt(2))
|BC|=2sqrt(3)/(sqrt(6)-sqrt(2))
|BC|=sqrt(3)(sqrt(6)+sqrt(2))/2
|BC|=(sqrt(6)+3sqrt(2))/2
From cosine law in triangle ABC calculate length |AC|
|AC|^2=3^2+|BC|^2-2*3*|BC|*cos(45)
|AC|^2= 9+1/4(6+18 + 6*2*sqrt(3))-2*3*(sqrt(6)+3sqrt(2))/2*sqrt(2)/2
|AC|^2 = 9 + 1/4(24+12sqrt(3)) - 6/4(2*sqrt(3)+2*3)
|AC|^2 = 9 + 6 + 3sqrt(3) - 3(3+sqrt(3))
|AC|^2 = 9 + 6 + 3sqrt(3) - 9 - 3sqrt(3)
|AC|^2 = 6
From cosine law in triangle ABC calculate cos(x)
|BC|^2 = |AB|^2+|AC|^2-2|AB||AC|cos(x)
1/4(6+18+2*3*2*sqrt(3)) = 9 + 6 - 2*3*sqrt(6)*cos(x)
1/4(24+12sqrt(3)) = 15 -6sqrt(6)cos(x)
6+3sqrt(3)=15-6sqrt(6)cos(x)
6sqrt(6)cos(x) = 15-6-3sqrt(3)
6sqrt(6)cos(x) = 9 - 3sqrt(3)
2sqrt(6)cos(x) = 3 - sqrt(3)
cos(x) = (3 - sqrt(3))/(2sqrt(6))
cos(x) = (3 - sqrt(3))sqrt(6)/12
cos(x) = (3sqrt(6)-3sqrt(2))/12
cos(x) = (sqrt(6) - sqrt(2))/4
x = 75 degrees
شكرا
يمكن حساب tanx نجد tanx=2+3^1/2اي °x=75
Alternate solution:
Drop perpendicular from C to point E on line AD. Let CE = h, and AE = w → ED = 2−w, EB = 3−w
In △CED, tan 60 = CE/ED → √ 3 = h/(2−w) → h = √ 3 (2−w)
In △CEB, tan 45 = CE/EB → 1 = h/(3−w) → h = 3−w
Therefore:
√ 3 (2−w) = 3−w → w = (2√3−3)/(√3−1) = (3−√3)/2
h = 3−w = (3+√3)/2
In △ACE:
tan x = CE/AE = h/w
= [(3+√3)/2] / [(3−√3)/2]
= (3+√3)/(3−√3)
= 2 + √3
= (1 + √3/2) / (1/2)
= (1 − cos 150) / sin 150 [(1−cos 2x)/sin 2x = tan x]
= tan 75
x = 75
Tgx=sqrt3/(2sqrt3-3)....x=75
Can u please explain how did you get angle APD as 90°? Thank u.
He used the inverse of 30-60-90 theorem. HE saw that PD is half of AD and one of the angle is 60, so he concludes that angle opposite to 2 is 90 degrees and remaining angle is 30 degrees.
He constructs it that way, he extends A such that meets CD at 90° at point P
PD is 1 unit and AD is 2 units. Sin 30 degrees is 1/2 Hence angle PAD is 30 degrees. and APD is 90 degrees
2:57 will a triangle with the side lengths 1 and 2 with a 60 degrees angle always have a 90 degrees angle?
I think that is a good question, or at least I asked myself the same question. Proving that a 30-60-90 triangle has a short leg half the length of the hypotenuse is straightforward. So I tried to see if the proof would work in reverse, and it does:
Given a line segment AB length 2 and another AC length 1 with a 60 degree angle between them at point A;
Connect a segment from B to C forming a triangle, and mark the midpoint of AB as point D, and connect D to C
In triangle ADC, AD (1/2 of 2)=AC (1) so triangle ADC is isosceles and angles ADC=ACD
Angle ADC + angle ACD + 60=180 so angles ADC and ACD also equal 60 so Triangle ACD is equilateral and AD=AC=DC=1
DB also equals 1 (1/2 of 2) and so also equals DC, so Triangle DCB is also isosceles
So angle DCB=angle DBC
And angle CDB is supplementary to angle CDA which is 60, so angle CDB=120 degrees
With the angles of triangle DCB summing to 180, angles DCB and DBC=30 degrees
And angle ACB= angle ACD+angle DCB=60+30=90 degrees
@@waheisel bạn giải thích thật chi tiết và dễ hiểu, cảm ơn bạn❤️❤️👍
I had exactly the same question, but I realized that knowing the angle and the length of the two adjacent sides would, indeed, resolve the triangle. He leaps to the conclusion, "... we can see ...", without explaining that this is a known special case that results in a right triangle. If you don't already know that 60 degrees with adjacent sides of x and 2x always gives that, then it seems like it's pulled out of nowhere. He could have gone into more detail.
Yes because it's similar to a 30-60-90 triangle with side lengths 1, 2, sqrt(3) by SAS similarity where 60 is the included angle.
@@johngeverett You've nailed it! 🙂
I can I need to revise all the geometrical theorems
Trick to solving these problems is to just draw a bunch of lines and apply mathematical rules until you find something.
Your method is more elegant but you don't need to be tricky to solve this:
Let h be the height of the triangle and z be the distance from point A to its base.
h = ztanx = (2 - z)tan60 = (3 - z)tan45
(2 - z)tan60 = (3 - z)tan45 --> z = (2 - sqrt3)/(1-sqrt3)
h = (3-z) --> h = [3(1 - sqrt3)-(2 - sqrt3)]/(1 - sqrt3)
h = ztanx --> tanx = h/z
--> {[3(1 - sqrt3)-(2 - sqrt3)]/(1 - sqrt3)}/[ (2 - sqrt3)/(1-sqrt3)]
--> tanx = 1/(2-3sqrt3) --> x = 75 deg.
Messy but straight forward.
I used the law of sine in five steps
There's another solution for that. I was able to answer it with different method.
Mọi thứ làm chi tiết nhưng điều cần chi tiết lại ko có. Như việc cần chứng minh tại sao tam giác APD lại vuông thì không chứng minh
can we solve this problem with PD=2 ?
Yes, but then AD = 4 🙂
👍
He starts by assuming that point P on line CD exists, but he never proves it.
in other words, maybe line CD is shorter then 1?
See my comments to CCM (above).
3:15 i didn t understand how you find APD =90 degree and PAD =30 degree !!!!
Draw a perpendicular on CD in Point P, draw AP, you've got it 🙂 (Due to the fact that we have a 60-degree-angle, the other one has to be a 30-degree-angle, in a right-triangle.)
Oh my god I did this on my own and got the answer
This problem is from Serbia junior national contest for 6th grade from last year (2021).
This problem actually appeared on a very old AMC test, I think from 2001 or something
Didnt know that, thanks for telling
Flase. Why APD=90°
60°
немного нудновато, но в целом мне понравилось! Привет из Антарктиды!
Angle A 75 degree
There’s another faster way to solve this problem, by quickly finding 4 RECTANGULAR TRIANGLES (no isoscele triangle) inside triangle ABC.
STEP A): we know that DB is the hypotenuse of a triangle having two 45° angles, and whose 2 cathetuses are both 0.707 long.
STEP B) We may find the length of another hypotenuse CD, of rectangular triangle having angles 75°, 15° and 90°, whose 2 cathetuses are 0.707 and 2.64 (cotg 15° . 0.707) = 2.73
STEP C) We may find another rectangular triangle CDE, having angles 30°, 60°, 90°, and whose side ED is = CD sin 30° = 2.73 . 0,5 = 1.365, and whose side CE is CD sin 60° = 2.73 . 0.86 = 2.36
STEP D) = We subtract the side ED from the length AD (2) and we get : 2 - 1.365 = 0.635 which is the side AE of another rectangular triangle CEA. And finally, having the sides CE (2.36) and AE (0.635), we easily find the hypotenuse AC, which is 2.44, and finally we find the angle x whose sine is simply CE/AC = 0.96 which is the sine of 75°, solution of our problem
That’s not faster
@@PegasusTenma1 Yes, it is ! You're in error. My method takes JUST 4 STEPS, whereas the other method they suggested takes 8 STEPS. Moreover, my solution needs JUST 1 method = PYTAGOREAN THEOREM, and it is much easier. And finally, I could summarize my solution in just 13 lines, and I could explain it in a video lasting no more than 3-4 minutes, much less than their 6. 41 minutes. Please, before criticizing think better!
@@Albert-ct6tt
Learn to accept when you’re wrong. The method you described used a lot of approximation and decimal places, as well as using trigonometry functions that cannot be evaluated without a calculator. Your method may contain “only” 4 steps but truthfully each one of those steps is longer than a single step presented in this video, it requires numerical calculations with decimals that also take more time than simple geometric construction.
I can end this question in merely 2 steps and less than a minute. Again, learn to accept when you’re wrong. Your method isn’t quick. It’s untidy and slow, and no one would use it.
@@PegasusTenma1Don't bother me, YOU ARE RIDICOLOUSLY WRONG! My method is using BASIC TRIGONOMETRY that even a normal 16-17 year old student should already know very well. Your problem is that YOU DON'T KNOW BASIC TRIGONOMETRY, and my method is "long" just for those who, like you, are totally ignorant in trigonometry. And every normal calculation through trigonometry needs decimals, don't you know it? Please, let's stop, I cannot waste my time with ignorant people like you.
Using sin and cosine laws, the answer will 75.000045 degrees
Cool
Thank you for your feedback! Cheers!
You are awesome, Mustafiz 😀
😅
Tavatır bir çözüm
The way you explain is complicated.
I hate that
The sum of angles on a straight line is 180 degree, so this could be solved mentally in 5 seconds.
60°
60°