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Motion In A Plane

Question
CBSEENPH11018335

If straight A with rightwards arrow on top times straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top times straight C with rightwards arrow on top and angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top is twice the angle between <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> then show that straight C over straight B equals 2 cos straight theta over 2 minus sec straight theta over 2 comma where straight theta is the angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top.

Solution

It is given that angle between straight A with rightwards arrow on top space and space straight B with rightwards arrow on top is straight theta. 
Therefore the angle between <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre> will be straight theta divided by 2.
Here,        straight A with rightwards arrow on top times straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top times straight C with rightwards arrow on top
∴         AB space cos space straight theta space equals space AC space cos left parenthesis straight theta divided by 2 right parenthesis
rightwards double arrow               straight C over straight B equals fraction numerator cosθ over denominator cos left parenthesis straight theta divided by 2 right parenthesis end fraction

space space space space space space equals fraction numerator 2 cos squared left parenthesis straight theta divided by 2 right parenthesis minus 1 over denominator cos left parenthesis straight theta divided by 2 right parenthesis end fraction

rightwards double arrow               straight C over straight B equals 2 cos space straight theta over 2 minus sec straight theta over 2.
Hence proved.