question_answer

In \[\Delta PQR,\] right-angled at Q, \[\text{PR}+\text{QR}=\text{25 cm}\] and \[\text{PQ}=\text{5 cm}\]. The value of tan P is:

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 \[QR=12\,cm\]

\[\therefore \,\,\,\,\,\,\,\,\,\tan P=\frac{QR}{PQ}=\frac{12}{5}\]