SOLUTION: An isosceles triangle ABC, in which AB = BC = 6√2 and AC = 12 is folded along the altitude BD so that planes ABD and BDC form a right dihedral angle. Find the angle between side

Algebra.Com
Question 1186931: An isosceles triangle ABC, in which AB = BC = 6√2 and AC = 12 is folded along the altitude BD so that planes ABD and BDC form a right dihedral angle. Find the angle between side AB and its new
position.

Found 2 solutions by yurtman, ikleyn:
Answer by yurtman(42)   (Show Source): You can put this solution on YOUR website!
**1. Find the Height (Altitude) of the Isosceles Triangle**
* **Use the Pythagorean Theorem:**
* In right triangle ABD (where D is the midpoint of AC):
* AB² = AD² + BD²
* (6√2)² = AD² + 6²
* 72 = AD² + 36
* AD² = 36
* AD = 6
**2. Visualize the Folding**
* When triangle ABC is folded along altitude BD, side AB rotates around point B.
* Imagine the original position of AB and its new position after folding. This creates a dihedral angle (the angle between the two planes).
**3. Determine the Angle Between AB and its New Position**
* The angle between AB and its new position is twice the angle between AB and the plane BDC.
* Let's call this angle θ.
* **Consider right triangle ABD:**
* tan(θ) = AD / BD = 6 / 6 = 1
* θ = arctan(1) = 45°
* **Angle between AB and its new position:** 2 * θ = 2 * 45° = 90°
**Therefore, the angle between side AB and its new position after folding is 90 degrees.**

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
An isosceles triangle ABC, in which AB = BC = 6√2 and AC = 12 is folded along the altitude BD
so that planes ABD and BDC form a right dihedral angle. Find the angle between side AB and its
new position.
~~~~~~~~~~~~~~~~~~~~


        The solution by the other tutor  (which,  I believe,  is an  AI  solution)  is  FATALLY  WRONG.
        I came to bring a correct solution.


In the isosceles triangle ABC

    AB^2 + BC^2 =  +  = 36*2 + 36*2 = 72 + 72 = 144

and
 
    AC^2 =  = 144.


It means that triangle ABC is a right-angled isosceles triangle with the right angle at B,
hypotenuse BC and the legs AB and BC.


Now imagine that the triangle ABC is folded along the altitude BD so the planes ABD and BDC form a right dihedral angle.
Let plane BDC remains in its original position and vertex A takes new position A' in space.
Then the segments A'D and CD form the right angle.


Consider triangle A'DC.  It is a right-angled triangle with the right angle D and the legs A'D and CD.


These legs have equal length  A'D = CD = 12/2 = 6, A'D= AD and AD and CD are equal halves of the base AC of the triangle ABC.


Therefore, the hypotenuse A'C of the triangle ADC has the length   =  = .


Thus the triangle A'BC  (after folding)  has congruent sides A'B, A'C and BC: the length of each of these sides is  .


Thus the triangle A'BC is an equilateral triangle and all its angles have the measure of 60°.

In particular, the angle A'BC is 60°.


The angle between AB and its new position A'B is the same 60°.  The proof is the same.

Solved.



RELATED QUESTIONS

in an acute Triangle ABC, an altitude AD is drawn. Find the area of triangle ABC if AB=... (answered by richwmiller)
In an acute triangle ABC,an altitude AD is drawn. Find the area of triangle ABC if AB=15, (answered by mananth)
Prove that ((AC)^2/(BC)^2)=(AD/BD) In right triangle ACB (C is the right angle) with CD... (answered by venugopalramana)
In triangle ABC, angle ABC = 90º, and point D lies on segment BC such that AD is an angle (answered by Edwin McCravy)
In right triangle abc, altitude ab. if ad = 8 cm and bd = 18 cm, find... (answered by jerryguo41)
what is the property of equality if AC = AB and AC = BD, then BD = AB +... (answered by richwmiller)
Angle C is the right angle. In right △ABC with right ∠C, AD bisects ∠BAC. Suppose... (answered by Solver92311)
ABC is an obtuse - angled triangle at A, AB = AC A=and intersects BC line segment at D (answered by Edwin McCravy)
in triangle abc in which DE is parallel to BC ,and intersects AB in D and AC in E . If... (answered by b_gulsum)