A) \[\frac{5}{12}\]
B) \[\frac{12}{5}\]
C) \[\frac{7}{5}\]
D) \[\frac{5}{7}\]
Correct Answer: B
Solution :
[b] Given, \[\angle Q=90{}^\circ \] and \[PQ=5cm\] |
\[PR+QR=25\,cm\] ..(1) |
In right \[\Delta PQR,\] |
\[P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}\] |
(by Pythagoras theorem) |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,P{{Q}^{2}}=P{{R}^{2}}-Q{{R}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{(5)}^{2}}=(PR+QR)\,\,(PR-QR)\] |
\[[{{a}^{2}}-{{b}^{2}}=(a+b)\,\,(a-b)]\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,25=25\cdot (PR-QR)\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,PR-QR=1\] |
From eqs. (1) + (2); |
\[2PR=26\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,PR=13\,cm\] and \[QR=12\,cm\] |
\[\therefore \,\,\,\,\,\,\,\,\,\tan P=\frac{QR}{PQ}=\frac{12}{5}\] |
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