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In triangle ABC, angles BAC and BCA = 80°, AB= BC ...

Created on: October 11, 2025

Answered using GPT-5 Thinking by Chat01

Question

In triangle ABC, angles BAC and BCA = 80°, AB= BC = 1/(2cos10°-√3). On side AB, there is a point E and on side BC, there is a point D, angle CAD = 50°, and angle ACE = 60°. Find ED

Python

python
import sympy as sp
sp.nsimplify(1/(2*sp.cos(sp.rad(10)) - sp.sqrt(3)))

1/(-sqrt(3) + 2*cos(pi/18))

Python

python
import sympy as sp
deg = sp.pi/180
b = 1/(2*sp.cos(10*deg) - sp.sqrt(3))
b_n = sp.N(b,50)
b_n


4.2093796195361659048780412331808185696886254299303

Python

python
import sympy as sp
deg = sp.pi/180
b = sp.nsimplify(1/(2*sp.cos(10*deg) - sp.sqrt(3)))
# Coordinates
A = sp.Matrix([b, 0])
B = sp.Matrix([0, 0])
C = sp.Matrix([b*sp.cos(20*deg), b*sp.sin(20*deg)])

# Define function to rotate a vector by angle theta
def rotate(v, theta):
    R = sp.Matrix([[sp.cos(theta), -sp.sin(theta)],
                   [sp.sin(theta),  sp.cos(theta)]])
    return R*v

# Compute E: line from C making angle 60° with CA
CA = A - C
# Two possible directions: rotate CA by +60° or -60°
u1 = rotate(CA, 60*deg)
u2 = rotate(CA, -60*deg)
# Intersect lines C + t*u with y=0
t1 = sp.solve(sp.Eq((C + sp.Symbol('t')*u1)[1], 0), sp.Symbol('t'))
t2 = sp.solve(sp.Eq((C + sp.Symbol('t')*u2)[1], 0), sp.Symbol('t'))
t1, t2


([2*sin(pi/9)/(-sqrt(3) + 2*cos(pi/18))],
 [2*sin(pi/9)/(-2*sin(2*pi/9) + sqrt(3))])

Python

python
t_sym = sp.Symbol('t')
u1 = sp.simplify(u1)
u2 = sp.simplify(u2)
t1 = t1[0]
t2 = t2[0]
E1 = sp.simplify(C + t1*u1)
E2 = sp.simplify(C + t2*u2)
sp.N(E1), sp.N(E2), sp.N(b)


(Matrix([
 [7.91104593312946],
 [               0]]),
 Matrix([
 [2.23976411351175],
 [               0]]),
 4.20937961953617)

Python

python
# Direction from A
AC = C - A
v1 = rotate(AC, 50*deg)   # rotate CCW from AC by +50
v2 = rotate(AC, -50*deg)  # rotate CW by -50

# Param lines: L1 = A + u*v1; L2 = A + u*v2
u = sp.Symbol('u', real=True)
s = sp.Symbol('s', real=True)
lineBC = B + s*(C - B)
# Solve for intersection with BC: find u and s s.t. A + u * v? = lineBC
sol1 = sp.solve(sp.Eq(A + u*v1, lineBC), [u, s], dict=True)
sol2 = sp.solve(sp.Eq(A + u*v2, lineBC), [u, s], dict=True)
sol1, sol2


([{s: -1 - 16*sin(pi/9)**4 + 16*sin(pi/9)**2,
   u: -40*sin(pi/9)**3 + 32*sin(pi/9)**5 + 8*sin(pi/9)}],
 [{s: 2 - 4*sin(pi/9)**2,
   u: -40*sin(pi/9)**3 + 32*sin(pi/9)**5 + 10*sin(pi/9)}])

Python

python
sol1_num = {k: sp.N(v) for k,v in sol1[0].items()}
sol2_num = {k: sp.N(v) for k,v in sol2[0].items()}
sol1_num, sol2_num


({s: 0.652703644666139, u: 1.28557521937308},
 {s: 1.53208888623796, u: 1.96961550602442})

Python

python
s_val = sol1[0][s]
D = sp.simplify(lineBC.subs(s, s_val))
E = sp.simplify(E2)
ED_len = sp.sqrt(sp.simplify(sp.sum((D - E)**2)))
sp.N(D), sp.N(E), sp.N(ED_len,50)

Thought

s_val = sol1[0][s] D = sp.simplify(lineBC.subs(s, s_val)) E = sp.simplify(E2) ED_len = sp.sqrt(sp.simplify(sp.sum((D - E)**2))) sp.N(D), sp.N(E), sp.N(ED_len,50)

Thought for 1m 1s

Python

python
ED_len = sp.sqrt(sp.simplify(((D - E).dot(D - E))))
sp.N(D), sp.N(E), sp.N(ED_len,50)


(Matrix([
 [ 2.58178425683742],
 [0.939692620785908]]),
 Matrix([
 [2.23976411351175],
 [               0]]),
 1.0000000000000000000000000000000000000000000000000)

Python

python
import sympy as sp
deg=sp.pi/180
x=sp.cos(20*deg)
y=sp.cos(40*deg)
S = sp.simplify(1/(4*x**2) + 1/(4*sp.cos(40*deg)**2) - 1/(2*sp.cos(40*deg)))
sp.N(S,50)


0.056436985952271091082210843128870364384253903149490

Python

python
b = 1/(2*sp.cos(10*deg) - sp.sqrt(3))
float(b**2 * S)


1.0

Python

python
delta = sp.simplify(S - (2*sp.cos(10*deg) - sp.sqrt(3))**2)
sp.simplify(delta)
sp.N(delta,50)


0.e-168

Thought

delta = sp.simplify(S - (2sp.cos(10deg) - sp.sqrt(3))**2) sp.simplify(delta) sp.N(delta,50)

0.e-168

Thought for 1m 6s

Answer

ED = 1

Sketch of solution

Let ∠A = ∠C = 80°, so ∠B = 20°. Put AB=BC=bAB=BC=b (the problem gives b=12cos⁡10∘−3b=\dfrac1{2\cos10^\circ-\sqrt3}).

  1. From ∠ACE=60∘\angle ACE=60^\circ and ∠ACB=80∘\angle ACB=80^\circ, we get ∠ECB=20∘\angle ECB=20^\circ. Also ∠EBC=∠ABC=20∘\angle EBC=\angle ABC=20^\circ.
    Hence in △BEC\triangle BEC the base angles at BB and CC are equal, so EB=ECEB=EC.
    By the sine rule:

    EBsin⁡20∘=BCsin⁡140∘⇒EB=EC=bsin⁡20∘sin⁡40∘=b2cos⁡20∘.\frac{EB}{\sin20^\circ}=\frac{BC}{\sin140^\circ}\quad\Rightarrow\quad EB=EC=\frac{b\sin20^\circ}{\sin40^\circ}=\frac{b}{2\cos20^\circ}.

  2. From ∠CAD=50∘\angle CAD=50^\circ and ∠CAB=80∘\angle CAB=80^\circ, we get ∠DAB=30∘\angle DAB=30^\circ. In △ABD\triangle ABD: ∠ABD=20∘, ∠BAD=30∘\angle ABD=20^\circ,\ \angle BAD=30^\circ.
    By the sine rule:

    BDsin⁡30∘=ABsin⁡130∘⇒BD=bsin⁡30∘sin⁡50∘=b2sin⁡50∘=b2cos⁡40∘.\frac{BD}{\sin30^\circ}=\frac{AB}{\sin130^\circ}\quad\Rightarrow\quad BD=\frac{b\sin30^\circ}{\sin50^\circ}=\frac{b}{2\sin50^\circ}=\frac{b}{2\cos40^\circ}.

  3. The angle between BEBE and BDBD at BB is ∠EBD=∠ABC=20∘\angle EBD=\angle ABC=20^\circ. Apply the cosine law in △EBD\triangle EBD:

    ED2=EB2+BD2−2⋅EB⋅BDcos⁡20∘=b2 ⁣(14cos⁡220∘+14cos⁡240∘−12cos⁡40∘).ED^2=EB^2+BD^2-2\cdot EB\cdot BD\cos20^\circ =b^2\!\left(\frac1{4\cos^2 20^\circ}+\frac1{4\cos^2 40^\circ}-\frac1{2\cos40^\circ}\right).

  4. Using sin⁡50∘=cos⁡40∘\sin50^\circ=\cos40^\circ, sin⁡40∘=2sin⁡20∘cos⁡20∘\sin40^\circ=2\sin20^\circ\cos20^\circ, and cos⁡20∘=2cos⁡210∘−1 \cos20^\circ=2\cos^2 10^\circ-1 (plus cos⁡3⋅20∘=12\cos3\cdot20^\circ=\frac12), the bracket simplifies to

    (2cos⁡10∘−3)2.\left(2\cos10^\circ-\sqrt3\right)^2.

    Therefore

    ED2=b2(2cos⁡10∘−3)2.ED^2=b^2\left(2\cos10^\circ-\sqrt3\right)^2.

With the given b=12cos⁡10∘−3b=\dfrac1{2\cos10^\circ-\sqrt3}, we get ED2=1ED^2=1, hence ED=1ED=\boxed{1}.

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