Find the angle X | How to Solve this Tricky Geometry problem Quickly

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.

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Find the angle X | How to Solve this Tricky Geometry problem Quickly

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Olympiad Mathematics
pre math
Po Shen Loh
Learn how to find the angle X
How to Solve this Tricky Geometry problem Quickly

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[johnbrennan]

Excellent method. Very interesting problem.

[pwmiles]

Very clever! I solved it by trigonometry, getting tan(x) = 1/(2-sqrt(3)). The video method is better though

[Ramkabharosa]

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!

[crmf]

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 🇧🇷

[WahranRai]

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

[panPetrff]

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 = <BAC = 150/2 = 75°

[Shubhagarwal_]

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

In triangle CEB,
TAN 45 = CE/ BE
1 = CE /( ED +BD)
1 = √3 ED / (ED + [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 👍

[Ramkabharosa]

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 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!
.

[Fred-yqfs]

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

[philipkudrna]

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!

[mva]

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.

[paullovett]

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.

[shadrana]

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

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

=(12+6sqrt3)/6=(2+sqrt3)
tanx=(2+sqrt3)
x=arctan(2+sqrt3)=75 degrees. This is our answer.

[jimlocke]

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 <CAE to the side adjacent is ((3 + √3)/2)/((3 - √3)/2) = (3 + √3)/(3 - √3). Those familiar with the 15°-75°-90° right triangle will recognize that this is the ratio of the long side to the short side. Therefore, <CAE = x = 75°.

(On a scientific calculator, you can compute arctan ((3 + √3)/(3 - √3)) and get 75° to within the calculator's roundoff error.)

[vanditseksaria]

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

[matilda]

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

[Ramkabharosa]

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.
.

[humester]

Even though it may be true. You never offered any proof that <PAD is 90 deg. Just because the sides are 1 and 2, it doesn't necessarily mean it is a right triangle.
I think you would first have to apply the cosine rule to find AP^2.
AP^2 = 1^2 +2^2 -2(1)(2)cos60=5-4(1/2)=3
Therefore, since 1^2 +2^2 = 3, it is a right triangle.

[sadeekmuhammadryan]

Perfect angle chasing. Your problems and solutions are getting better day by day. Keep going! ❤️🇧🇩

[mander]

I was actually able to do this in my head.