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Some Applications Of Trigonometry

Question
CBSEENMA10007926
Wired Faculty App

If ∠ZB and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.

Solution
Let us consider two right angle triangles right angled at C and R respectively.
sinB space equals space AC over AB
Now,
  
and     <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) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>
   But sin space straight B space equals space sin space straight Q space left parenthesis given right parenthesis
∴        AC over AB equals PR over PQ
rightwards double arrow space space space space space space space space AC over PR equals AB over PQ
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In ΔABC,
Using Pythagoras theorem, we have
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In increment PQR comma
Using Pythagoras theorem, we have
                      QR space equals space square root of PQ squared minus PR squared end root
So,           BC over QR equals fraction numerator straight K square root of PQ squared minus PR squared end root over denominator square root of PQ squared minus PR squared end root end fraction equals straight K space space space... left parenthesis ii right parenthesis
Comparing (i) and (ii), we get
                 AC over PR equals AB over PQ equals BC over QR
So, by using SSS similar condition
∴          space space increment ACB space tilde space increment PRQ
                angle straight B space equals space angle straight Q.
   

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