Question 567470
Each hour mark on the clock is {{{ 5/60 = 1/12 }}} of
a {{{ 360 }}} degree circle apart. That is {{{ (1/12)*360 = 30 }}} degrees
At 7 pm, the minute hand and hour hand are
{{{ 7*30 = 210 }}} degrees apart
The minute hand is moving at the rate of {{{ 360 / 60 = 6 }}} degrees/min
The hour hand is moving at the rate of {{{ 30/60 = 1/2 }}} degree/min
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Let {{{ d[1] }}} = the degrees that the minute hand travels until it makes
an angle of {{{ 90 }}} degrees with the hour hand
Let {{{ d[2] }}} = the degrees that the hour hand travels until it makes
an angle of {{{ 90 }}} degrees with the minute hand
Let {{{ t }}} = time in minutes until they are {{{ 90 }}} degrees apart
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(1) {{{ 210 + d[2] - d[1] = 90 }}}
(2) {{{ 6t  = d[1] }}}
(3) {{{ (1/2)*t = d[2] }}}
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This is 3 equations and 3 unknowns, so it's solvable
Substitute (2) and (3) into (1)
(1) {{{ 210 + (1/2)*t - 6t = 90 }}}
(1) {{{ 420 + t - 12t = 180 }}}
(1) {{{ -11t = 180 - 420 }}}
(1) {{{ 11t = 240 }}}
(1) {{{ t = 21.818 }}} min
{{{ .818*60 = 49.09 }}} sec
The hands will be 90 degrees apart at 21 min and 49 sec past 7 PM
hope I got it