Question-187893

Question Number 187893 by Rupesh123 last updated on 23/Feb/23

Answered by a.lgnaoui last updated on 23/Feb/23

△ A B D \frac{\sin 53}{A D} = \frac{\sin (14 + x)}{A B} = \frac{\sin (14 + 53 + x)}{2 k}

2 k \sin 53 = A D \sin (67 + x) (1)

△ A C D \frac{\sin 14}{k} = \frac{\sin x}{A D}

k \sin x = A D \sin 14 (2)

\frac{(1)}{(2)} \Rightarrow \frac{2 \sin 53}{\sin x} = \frac{\sin (67 + x)}{\sin 14}

2 \sin 53 \times \sin 14 = \sin x \times \sin (67 + x)

= \sin^{2} x \times \cos 67 + \sin x \cos x \sin 67

= \cos 67 \sin^{2} x + \sin 67 \sin x \sqrt{1 - \sin^{2} x}

1 - \sin^{2} x = {(2 \sin 53 \sin 14 - \cos 67 \sin^{2} x)}^{2} \times \frac{1}{\sin^{2} 67 \sin^{2} x}

p o s i n s \sin x = z

\sin 53 = a \sin 14 = b \cos 67 = c

\sin 67 = d

d^{2} z^{2} (1 - z^{2}) = {(2 a b - {c z}^{2})}^{2}

d^{2} z^{4} - 2 d^{2} z^{2} + d^{2} z^{2} = 4 a^{2} b^{2} + c^{2} z^{4} - 4 {a b c z}^{2}

z^{2} = t

(d^{2} - c^{2}) t^{2} + (4 a b c - d^{2}) t - 4 a^{2} b^{2} = 0

t^{2} + \frac{4 a b c - d^{2}}{d^{2} - c^{2}} t - \frac{4 a^{2} b^{2}}{d^{2} - c^{2}} = 0

{(t + \frac{4 a b c - d^{2}}{2 (d^{2} - c^{2})})}^{2} - (\frac{4 a^{2} b^{2} (d^{2} - c^{2})}{4 {(d^{2} - c^{2})}^{2}} + \frac{(4 a b c - d^{2})}{4 {(d^{2} - c^{2})}^{2}}) = 0

{\begin{cases} t + \frac{4 a b c - d^{2}}{2 (d^{2} - c^{2})} \pm \frac{\sqrt{4 a^{2} b^{2} (d^{2} - c^{2}) + 4 a b c - d^{2}}}{2 (d^{2} - c^{2})} = 0 \\ t = \sin^{2} x x = a r c \sin \sqrt{t .} \end{cases}

(a s u i v r e) \dots \dots . .

Answered by mr W last updated on 23/Feb/23

m = “ {m e d i a n}^{″}

\frac{m}{k} = \frac{\sin x}{\sin 14 °} \dots (i)

\frac{2 k}{m} = \frac{\sin (53 ° + 14 ° + x)}{\sin 53 °} \dots (i i)

(i) \times (i i) :

2 = \frac{\sin x}{\sin 14 °} \times \frac{\sin (53 ° + 14 ° + x)}{\sin 53 °}

2 \sin 14 ° \sin 53 ° = \sin^{2} x \cos 67 ° + \sin x \cos x \sin 67 °

4 \sin 14 ° \sin 53 ° = (1 - \cos 2 x) \cos 67 ° + \sin 2 x \sin 67 °

\cos 67 ° - 4 \sin 14 ° \sin 53 ° = \cos (2 x + 67 °)

x = \frac{\cos^{- 1} (\cos 67 ° - 4 \sin 14 ° \sin 53 °) - 67 °}{2}

\approx 22.732 °

Commented by Rupesh123 last updated on 23/Feb/23

Very nice solution!