Question-11206

Question Number 11206 by uni last updated on 16/Mar/17

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Mar/17

A D . A C = 6 \times 24 = 144

{A D}^{2} + {A C}^{2} = 24^{2} = 576

{(A D + A C)}^{2} = 576 + 2 \times 144 = 6 \times 144 \Rightarrow

A D + A C = 12 \sqrt{6}

t^{2} - 12 \sqrt{6} t + 144 = 0 \Rightarrow

t = \frac{12 \sqrt{6} \pm \sqrt{6 \times 144 - 4 \times 144}}{2} = \frac{12 \sqrt{6} \pm 12 \sqrt{2}}{2}

t_{1} = A C = 6 (\sqrt{6} + \sqrt{2}), t_{2} = A D = 6 (\sqrt{6} - \sqrt{2})

\sin C = \frac{A D}{24} \Rightarrow \sin C = \frac{6 (\sqrt{6} - \sqrt{2})}{24} \Rightarrow C = 15^{°}

∡ C D A = 75^{°} \Rightarrow \frac{B D}{s i n 15^{°}} = \frac{x}{s i n (180 - 75)}

B D = x \frac{s i n 15}{s i n 75} = x . t g 15

x . A C . s i n (90 + 15) = (B D + 24) . 6 \Rightarrow

x . 6 (\sqrt{6} + \sqrt{2}) c o s 15 = (x t g 15 + 24) \times 6

x (\sqrt{6} + \sqrt{2}) (\frac{\sqrt{6} + \sqrt{2}}{4}) = x (2 - \sqrt{3}) + 24 \Rightarrow

x (2 + \sqrt{3} - 2 + \sqrt{3}) = 24 \Rightarrow x = \frac{24}{2 \sqrt{3}} = 4 \sqrt{3}

t_{1} = A C = 6 (\sqrt{6} - \sqrt{2}), t_{2} = A D = 6 (\sqrt{6} + \sqrt{2})

\Rightarrow ∡ C = 75, ∡ C D A = 15

\frac{B D}{s i n 75} = \frac{x}{s i n (180 - 15)} \Rightarrow B D = x c o t g 15

x . A C . s i n (90 + 15) = (B D + 24) \times 6 \Rightarrow

x . 6 (\sqrt{6} - \sqrt{2}) . c o s 15 = (x . c o t g 15 + 24) \times 6

x (\sqrt{6} - \sqrt{2}) (\frac{\sqrt{6} + \sqrt{2}}{4}) = x (2 + \sqrt{3}) + 24

x (1) = x (2 + \sqrt{3}) + 24 \dots \dots . i m p o s s i b l e .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

D H . C H = 6^{2} = 36, D H + C H = 24

C H = \frac{24 \pm \sqrt{24 \times 24 - 4 \times 36}}{2} = 12 \pm 6 \sqrt{3}

t g C = \frac{12 \pm 6 \sqrt{3}}{6} = 2 \pm \sqrt{3} \Rightarrow ∡ C = 15, 75

\Rightarrow ∡ C = 15, 75 \Rightarrow ∡ D A H = 15, 75 (i m p o s s i b l e)

\Rightarrow ∡ B A H = 30, B H = \frac{1}{2} A B = \frac{1}{2} x

B H = \frac{1}{2} x \Rightarrow x^{2} = {(\frac{1}{2} x)}^{2} + 6^{2} \Rightarrow

\frac{3}{4} x^{2} = 36 \Rightarrow x^{2} = 48 \Rightarrow x = 4 \sqrt{3}

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

x . A C . s i n (90 + 15) = 6 \times (B D + 24) . . (i)

\frac{B D}{s i n 15} = \frac{x}{s i n (B D A)} = \frac{x}{s i n (90 + C)} = \frac{x}{c o s C}

A D = 24 s i n C, A C = 24 c o s C

A C . A D = 6 \times 24 \Rightarrow 24 c o s C . 24 s i n C = 6 \times 24

s i n 2 C = \frac{1}{2} \Rightarrow 2 C = 30 \Rightarrow ∡ C = 15

\frac{B D}{s i n 15} = \frac{x}{c o s 15} \Rightarrow B D = x t g 15

(i) \Rightarrow x . 24 c o s 15 . c o s 15 = 6 . (x t g 15 + 24)

4 x . {c o s}^{2} 15 = x t g 15 + 24

x (4 \times {(\frac{\sqrt{6} + \sqrt{2}}{4})}^{2} - (2 - \sqrt{3})) = 24 \Rightarrow

x (2 + \sqrt{3} - 2 + \sqrt{3}) = 24 \Rightarrow x = \frac{24}{2 \sqrt{3}} = 4 \sqrt{3}