Question

please solve, a question from indian insitute of technology (IIT JEE 2014) exams

Answer 1

   This kind of problems are called Langley's Hardest Easy Math problems. They could be solved using Geometry by constructing Isosceles, equilateral triangles and using geometry identities in the triangles.

  Since it is for grade XII, we can use also trigonometry and coordinate geometry. Here is a solution. If you are allowed use of calculator, it is much simpler.

   We deduce quickly that AB = BC (= b say), as ∠B = ∠C = 80°.  Let the 
Coordinates of B(0, 0), A(b Cos80°, b Sin 80°), C(2b Cos80°, 0).

Equation of line AB:   y = x tan80°   ---(1)
Equation of line CD:  (y - 0)/(x - 2b Cos80°) = - tan60°  ----(2)
  
So for point D:  (1) = (2).
     x = 2b cos80° Tan60° / (Tan80° + tan60°)
        = 2 b Cos² 80° Sin60° /Sin140° = 2b Cos²80° Sin60° / Sin40°
     y = 2b Sin80° cos80° sin60°/sin40° = b sin160° sin60° / sin40°
        = b Sin20° Sin60° / Sin40°
   D = (2b Cos²80° Sin60° /Sin40° , b Sin20° Sin60 /SIn40°)   ---(3)

Now equation of line BE :  y = x Tan70°   ---(4)
   equation of line AC:  (y - 0)/(x - 2b Cos80°) = - tan80°
        or, AC:  y = -x Tan80°+ 2b Sin80°    ---(5)

Point E:  from (4) & (5):    x = 2b Sin80° / (Tan70° +Tan80°)
        x = 2b Sin 80° Cos80° Cos70° / Sin(70°+80°)
           = b Sin160 Cos70 / Sin 150° = b Sin20° Sin20° / SIn30°
           = 2b Sin² 20° .
        y = 2b Sin²20° Tan70° = 2b Sin²20° Cot20° 
           = 2b Sin20° Cos20° = b Sin40°
      E(2b Sin°20°, b Sin40°)

Slope of DE = m = (y2-y1)/(x2 - x1)

$m = \frac{bSin40-\frac{bSin20 \: Sin60}{Sin40}}{2bSin^220-\frac{2bCos^280 \: Sin60}{Sin40}}\\\\=\frac{Sin^240-Sin20 \: sin60}{2Sin^220 \: Sin40-2Cos^280 \: Sin60}\\\\=\frac{\frac{1}{2}(1-Cos80)-\frac{1}{2}(Cos40-Cos80)}{Sin20(Cos20-Cos60)-Cos80(Sin40-Sin20)}\\\\=\frac{(1-Cos40)/2}{Sin20 (Cos20-Cos60+Cos80)-Cos80*Sin40}\\\\=\frac{Sin^2 20}{Sin20(Cos20-Cos60+Cos80-Cos80*2Cos20)}\\\\=\frac{Sin 20}{Cos20-Cos60+Cos80-Cos80*2Cos20}\\\\=\frac{Sin 20}{Cos20-Cos60+Cos80-Cos100-Cos60}$

$m=\frac{Sin 20}{(1- 2 Sin^2 10) + 2 Sin10 - 1}\\\\==\frac{2 Sin10 \: Cos10}{-2 Sin^2 10 + 2 Sin10}\\\\=\frac{Cos10}{1-Sin10}=\frac{cos^2 5- sin^2 5}{(cos 5 - sin 5)^2}=\frac{cos 5 + sin5}{cos5 - sin5}\\\\=\frac{1+tan5}{1-tan5}=Tan(45^0+5^0)\\\\=Tan50^0$

Angle made by DE with horizontal = 50°.

At point E:  y = 180° - 80° - 50° - 30° = 20°

From ΔDOE,      angle x = 180° - 50° - 20° = 110°

Answer = x = 110°

Answer 2

Heya user,

I'll be using Complete Trigonometry to solve your problem + A trick + Sine Rule + the fact that sin A = cos ( 90 - A ) Hehe

--> Let AB = AC = x
==> BC = x sin 20° / sin 80° <-- In ABC

Now, CE = BC sin 70° / sin 30° <--- In Triangle BCE
-----> OC = BC sin 70° / sin 50° <--- In Triangle OBC
-----> AD = CD = BC sin 80° / sin 40° <--- In Triangle BDC
-----> BE = AE sin 20° / sin 10° <--- In triangle ABE <--- (i)
-----> BE = BC sin 80° / sin 30° <--- In triangle BCE <--- (ii)

From (i) and (ii) AE = BC sin 80° sin 10° / ( sin 30° sin 20° )

Now, CE / OC = sin 50° / sin 30° = 2 sin50°
Also, AD / AE = [ sin 30° sin 20° ] / [ sin 40° sin 10° ]

                        = [ 2 sin 10° cos 10° ] / [ 2 sin 40° sin 10° ]

                        = [ cos 10° ] / [ sin 40 ]° = sin 80° / sin 40°

                        = 2 cos  40° = 2 sin 50°

Now, since, CE / OC = AD / AE and Angle C = Angle A,

                        Triangle ECO is similar to Triangle DAE

=> AED = COE = 130° and ADE = CEO = 30°

Hence, x = 110° .... Hope you find it EASY too !!!