Find the Length X | An Amazing Geometry Question

[@Mycroft616]

I was with you until the end. Once I saw it was a 30-60-90 Right Triangle, I knew x + 1 was equal to x × 3^(1/2) and solved for x that way. I got the same answer, but was not bogged down by the negative root.

[@saidtakadda6799]

Hello, an other methode using al-khashi formula, and it can work for any angle wa have (2x)²=(x+1)²+x²-2*(x+1)*x*cos(3teta) and we have too x²=(x+1)²+(2x)²-2*(x+1)*2x*cos(teta) and we add the two equation and simplify x² and 4x² we have 2*( (x+1)² - (x+1)*x*cos(3teta) - 2*(x+1)*x*cos(teta))=0 and we have x+1>0 so the equation become x+1 - x *(cos(3teta)+2cos(teta)) =0 so x= 1/(cos(3teta) + 2cos(teta) - 1) in the case of test=30 which is our case cos(teta) =squart(3)/2 ans cos(3teta)=0 x= 1/squart(3)-1 = squart(3)+1/2

[@sumankundu93]

Not the most elegant apporach but one can use the formula sinA/a=sinB/b=sinC/c

[@احمدمحمد-ت8ط9ج]

تمرين جميل جيد. رسم واضح مرتب . شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .

[@MarieAnne.]

Nice geometric proof. I did this using law of sines: sin 3θ / (2x) = sin θ / x sin 3θ / sin θ = 2x / x (3 sin θ − 4 sin³θ) / sin θ = 2 3 − 4 sin²θ = 2 4 sin²θ = 1 sin θ = 1/2 [Note: we ignore negative root, since 0 < θ+3θ < 180 → 0 < θ < 45 → sin θ > 0] θ = 30° → 3θ = 90° → Third angle = 60° Using law of sines again we get: sin 90° / (2x) = sin 60° / (x + 1) 1 / (2x) = (√3/2) / (x + 1) 2x (√3/2) = x + 1 √3 x − x = 1 x (√3 − 1) = 1 x = 1/(√3−1) = (1 + √3) / 2

[@honestadministrator]

Herein 2x/ x =sin ( 3 theta) / sin (theta) =[ sin (2 theta) cos (theta)+ cos (2 theta) sin (theta) ] / sin (theta) = 2 cos^2 (theta) + cos (2 theta) = 1+ cos (2 theta) + cos (2 theta) = 1 + 2 cos ( 2 theta) [ Again 0 < 3 theta < π ] This implies cos (2 theta) = 1/2 = cos( π/3) So only feasible solution is 2 theta = π/3 i.e. 3 theta = π/2 Hereby AC = √ (( 2x) ^2 - x^2) = x √3 Therefore x √3 = x +1 or x = 1/( √3 -1) = (√3 +1) /2.

[@honestadministrator]

sin( 3 theta) /(2x) = sin(theta) / x cos(2 theta) + 2 cos^2 (theta) = 2 1 + 2 cos (2 theta) = 2 cos(2 theta) = 1/2 = π/3 theta= π/6 Hereby 3 theta = π/2 (2 x) ^2 = (x) ^2 + ( x+1)^2 √3 x = x + 1 x = 1/( √3 -1) = ( √3 +1 ) /2

[@gauravjain1974]

you can also solve this problem by using trignometry. first, use the laws of sines. sin 3(theta) /2x = sin(theta)/x. multiply both sides by x. now, the equation becomes sin 3(theta)/2 = sin(theta). now, sin 3(theta) = 3(sin(theta)) - 4(sin(theta))^3. then, form the equation sin 3(theta)/sin(theta) = 2. since we know that sin 3(theta) = 3(sin(theta)) - 4(sin(theta))^3, we can substitute that in and cancel out sin(theta) from both numerator and denominator. now we have the equation 3 - 4(sin(theta))^2 = 2. 4(sin(theta))^2 becomes 1. (sin(theta))^2 becomes 1/4. sin(theta) becomes 1/2, or -1/2. now, since arcsin(1/2) is -pi/6, that cant be the answer, since it is negative. so the only root of sin(theta) will be 1/2. and arcsin of 1/2 is 30. so, theta becomes 30 degrees, and 3*theta is 90 degrees. now, using the pythagoreas theorem, we can make the equation x^2 + (x+1)^2 = (2x)^2, which then becomes 2x^2 + 2x + 1= 4x^2. now the equation becomes 2x^2 - 2x - 1 = 0. so, x becomes (2 + sqrt(12))/4, which is equal to (1+ sqrt(3))/2. the other root of the equation wont be possible, since it is negative.

[@sumit180288]

Simply apply sine rule and its a piece of cake

[@alinayfeh4961]

In ABD two lengths hypotunese AB -AD, draw height AH The Area ADC=ADB, AHx/2, sin((2theta))/x=((thet))/x, 2((theta))=60⁰ ((Thta))=30⁰, x=(1+√(3)/2

[@alexandroyassuhiro6514]

There is the Egyptian triangle . If I have angles 30, 60, and 90. So the longer side is y, the shorter side is y/2, and the other side is y√3/2. 2X = Y√3/4. =. X= y√3/4, X+1 = 1+y√3/4. If X = 2 + y√3/4, X = 1+√3/2

[@spafon7799]

When I saw one side is twice the length of another side my immediate thought is, is this a 30-60-90? And then we see that one angle is 3 times another and it definitely is a 30-60-90. So then you know from the relative lengths of the sides of a 30-60-90 that x+1=sqrt(3)*x. Thus x*(sqrt(3)-1)=1, x=1/(sqrt(3)-1) Multiply numerator and denominator by sqrt(3)+1 and you get x=(sqrt(3)+1)/2. The only argument one can make to question this solution is what is the proof that it's a 30-60-90. But once you calculate these lengths of the sides of this triangle, you know that since these are the relative lengths of a 30-60-90 that conversely something that has these relative lengths has to be a 30-60-90.

[@guyhoghton399]

Marvellous! Clever idea to split angle  that way. There is also a trivial solution. If you let angle B̂ increase to 180° the apex A will lie to the left of point B making ΔABC collapse to a degenerate triangle forming the straight line segment: A -- x -- B ---- 2x ---- C Now angle Ĉ = θ = 0° and angle  = 3θ = 0°. Since AC = x + 1 we have 3x = x + 1 ∴ x = ½. This comes from applying the sine rule. We have: sinθ/x = sin3θ/2x ∴ sinθ = ½sin3θ ⇒ sinθ(4sin²θ - 1) = 0 Also 0 ≤ 3θ ≤ 180° ⇒ 0 ≤ θ ≤ 60° The first factor gives the trivial solution θ = 0°, the second θ = 30°.

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين

[@MathBooster]

Thank you 🙂

[@krishnamoyghosh6047]

It can be very simple solution by applying sin and aide theorem. Here sin of theta comes 30 degree.

[@grin2groin._.624]

Sine rule does it super fast. Got it in 2 mins

[@kylekatarn1986]

I've used the theorem of sine in order to discover that theta is 30°

[@idvbane8580]

Math is ❤❤❤❤

[@idvbane8580]

Trigonometry 🥰🥰😍

[@idvbane8580]

V gd explain 🥰🥰🥰🥰

[@MathBooster]

Thanks 😊

[@advaykumar9726]

Why didn't you use sine rule?

[@MathBooster]

You can also use sine rule to solve the problem.

[@yakupbuyankara5903]

X=((3^(1/2)+1)/2.

[@idvbane8580]

🥰🥰🥰

[@idvbane8580]

Math 🥰🥰

[@idvbane8580]

Sir😍🥰🥰🥰

[@idvbane8580]

Algebra🥰🥰

[@nurali45303]

İ did this easier vay sin3a=sina+sin2a

[@idvbane8580]

V gd explain 🥰🥰🥰

[@MathBooster]

Thanks 😊

[@idvbane8580]

🥰🥰🥰