1. ## Trigonometry

A man begin the trip when the hands of the clock are attached between 8 and 9 hours; the journey ends when the child pointer is between 14 and 15 hours and the pointer larger is 180° the other. How long the trip lasted ?

2. Hello, Apprentice123!

I believe I've translated your problem to English . . .

A man begin the trip when the hands of the clock are together between 8:00 and 9:00.
The journey ends when the hour hand is between 14:00 and 15:00 hours
and the minute hand is diametrically opposite. .How long did the trip last?
The minute hand moves at 6 degress/minute.
. . In $t$ minutes, $M$ moves: $6t$ degrees.

The hour hand moves at ½ degrees/minute.
. . In $t$ minutes, $H$ moves: $\tfrac{1}{2}t$ degrees.

At 8:00, the minute hand is at 0°; the hour hand is at 240°.
. . In $t$ minutes, $M$ is at $6t$ degrees . . . and $H$ is at $240 + \tfrac{1}{2}t$ degrees.
The hands are together: . $6t \:=\:\tfrac{1}{2}t + 240 \quad\Rightarrow\quad t \:=\:\frac{480}{11} \:=\:43\tfrac{7}{11}$
. . The trip started at: . $8\!:\!43\tfrac{7}{11}$ a.m.

At 2:00, $M$ is at 0°; $H$ is at 60°.
In $t$ minutes, $M$ is at $6t$ degrees . . . and $H$ is at $60 + \tfrac{1}{2}t$ degrees.
The hands are 180° apart: . $6t \:=\:(60+\tfrac{1}{2}t) + 180 \quad\Rightarrow\quad t \:=\:\frac{480}{11} \:=\:43\tfrac{7}{11}$
. . The trip ended at: . $2\!:\!43\tfrac{7}{11}$ p.m.

Therefore, the trip took exactly 6 hours.

3. Thank you, sorry my english