A Nice Geometry Problem | You should know this trick !! 

Extend AD to F such that ∆ABF is an Equilateral triangle. ∆BDF is Similar to ∆ADC. Hence, DF = (3/2)*AD AF = AD+DF = (5/2)AD= 5X/√3 Since AF = 15, X = 3√3

At 4:35, we have found that length AD = 6.On the same screen, an expression for the area of ΔACD is given as (1/2)(AC)(AD)sin(60°) which equals (1/2)(10)(6)((√3)/2) = 15√3. If AC is treated as the base of ΔACD and DE as its height, its area = (1/2)(10)(X) = 5X. Equating the 2 expressions for area of ΔACD gives 5X = 15√3 or X = 3√3, as Math Booster also found.

Interesting solution. Another solution, without computing AD: As BAD = CAD, ΔABD : ΔACD = (1/2 * AB * AD * sin BAD) : (1/2 * AC * AD * sin CAD) = AB : AC, so ΔACD = ΔABC * AC / (AB + AC) ΔABC = (1/2) * 10 * 15 * sin 120 = (75/2)√3, so ΔACD = (75/2)√3 * 10 / 25 = 15√3. ΔACD = (1/2) * 10 x = 5x, so 5x = 15√3, which yields x = 3√3

Nice! φ = 30°; ∆ ABC → AB = 15; AC = 10 = AE + CE; BC = BD + CD = a + b; DAB = CAD = 2φ = EAD → sin⁡(DEA) = 1; DE = x = ? 15/a = 10/b → b = 2a/3; sin⁡(4φ) = sin⁡(6φ - 4φ) = sin⁡(2φ) = √3/2 → cos⁡(4φ) = -cos⁡(6φ - 4φ) = -cos⁡(2φ) = -1/2; ∆ ABC → (BC)^2 = 225 + 100 - 2(15)(10)cos⁡(4φ) = 475 → BC = 5√19 = 5a/3 → a = 3√19 → b = 2√19 → AD = √((15)(10) - (3√19)(2√19)) = 6 → AE = 3 → x = 3√3 again, with cos(α) = ? and cos⁡(β) = ? φ = 30°; ∆ ABC → AB = 15; AC = 10 = AE + CE; BC = BD + CD = a + b; DAB = CAD = 2φ DE = x; EAD = 2φ; DEA = 3φ → ADE = φ → AE = x√3/3 → AD = 2x√3/3; ABD = α; DCA = β 15/a = 10/b → b = 2a/3 → BC = 5a/3 → ∆ ABC → CAB = 4φ → cos⁡(4φ) = -cos⁡(6φ - 4φ) = -cos⁡(2φ) = -sin⁡(φ) = -1/2 → (5a/3)^2 = 225 + 100 - 2(15)(10)cos⁡(4φ) → a = 3√19 → b = 2√19 → ∆ ABD → 9(19) = 225 + 3x^2 - 15x√3 → x1,x2 = 3√3; 2√3 → x = 3√3 → AD = 6 → x√3/3 = 3 → 36 = 225 + 9(19) - 2(15)(3√19)cos⁡(α) → cos⁡(α) = 4√19/19 36 = 100 + 4(19) - 2(10)(2√19)cos⁡(β) → cos⁡(β) = 7√19/38 → α + β = ? cos⁡(α) = 4√19/19 → sin⁡(α) = √(1 - cos^2(α)) = (√3)(√19/19) cos⁡(β) = 7√19/38 → sin⁡(β) = (3/2)(1/19)(√3)(√19) → sin⁡(α + β) = sin⁡(α)cos⁡(β) + sin⁡(β)cos⁡(α) = (√3/19)(√19)(7/2)(√19/19) + (3/2)(√3)(√19/19)(4√19/19) = √3/2 = sin⁡(2φ) → α + β = 60°

Sir i have doubt i did in 2 methods Initially i used cosine law tonfind the unknown side first method: I used angle bisecor theorem to find DC Then i found AE by ising tan60 w.r.t. X It came as AE=X/√3 Then i found EC =(10-X/√3) I used Pythagoras theorem in triangle DEC (10-x/√3) ²+x²=(2√19) ² I got x as 3√3 and 2√3 Why should we reject 2√3 sir can you plz say Second method After using angle bisector theorem i used stewarts theorem ro find AD i got as 6 then i found x i got only 3√3 Why we should reject 2√3 sir

3 √3

X=3×(3^(1/2)).

X=3√3 Cop teorema dei seni