Question

In ΔPQR, right-anlged at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

Answer 1

Answer:

⇒PQ=5cm

⇒PR+QR=25cm

⇒PR=25−QR

Now, In △PQR

⇒(PR) 2

=PQ 2 +QR 2

⇒(25−QR) 2 =5 2 +QR 2

⇒625+QR 2 −50QR=25+QR 2

⇒50QR=600

⇒QR=12cm

⇒PR=25−12=13cm

.

$\bf: ⇒50QR=600</h3><h3>⇒QR=12cm</h3><h3>⇒PR=25−12=13cm$

∴sinP= PR QR = 13 12 ,cosP= PR PP = 135 ,tanP= PQ QR = 512

Hence, the answers are sinP= 1312 ,cosP= 135 ,tanP= 512 .

$\__________$

Answer 2

${ \mathbb{ \colorbox{black} { \boxed{ \boxed{ \colorbox{lime} {-:Answer:-}}}}}}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{red}{Given::}}}}}}$

$\pink{➠} \sf{ \triangle PQR \: is \: a \: right - angled \: triangle. }$

$\pink{➠} \sf{ \angle Q \: is \: a \: right \: angle \: of \: triangle. }$

$\pink{➠} \sf{PR+QR=25cm }$

$\pink{➠} \sf{PQ=5cm }$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{magenta}{To \: find::}}}}}}$

$\pink{➠} \sf{sin \: P }$

$\pink{➠} \sf{cos \: P }$

$\pink{➠} \sf{tan \: P }$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{black}{Formula \: used::}}}}}}$

$\pink{➠} \sf{sin \: P = \frac{side \: opposite \: to \: \angle P}{hypotenuse} }$

$\pink{➠} \sf{cos \: P = \frac{side \: adjacent \: to \: \angle P}{hypotenuse} }$

$\pink{➠} \sf{tan \: P = \frac{side \: opposite\: to \: \angle P}{side \: adjacent \: to \: \angle P} }$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{blue}{According \: to \: Question::}}}}}}$

$\bf{Let's \: start \: now!!! }$

$\sf{Let \: QR={ \bf \: x } \: cm}$

$: :\implies\sf{PR+QR=25cm }$

$: :\implies\sf{PR = 25 \: cm - QR }$

$: :\implies\sf{PR = 25 \: cm -{ \bf x } } \: \: \: \: \: \: \: \: \: \: ...(i)$

$\sf{In \: \triangle PQR, }$

${: : \implies \sf{(PR)}^{2} = {(PQ)}^{2} + {(QR)}^{2} \: \: \: (By \: pythagoras \: theorem)}$

${: : \implies \sf{(25 - \bf x)}^{2} = {(5)}^{2} + {( \bf \: x)}^{2}}$

${: : \implies \sf{ {(25)}^{2} - \bf (x)}^{2} - 2 \times 25 \times {\bf \: x} = {(5)}^{2} + {( \bf \: x)}^{2}}$

${: : \implies \sf{625 + { \bf \: x}^{2} - 50 {\bf \: x} = 25 + { \bf \: x}^{2} }}$

${: : \implies \sf{625 + { \bf \: x}^{2} - 50 {\bf \: x} - 25 - { \bf \: x}^{2} = 0 }}$

${: : \implies \sf{{ \bf \: x}^{2} - {\bf \: {x}^{2} } - 50 {\bf \: x} + 625 - 25= 0 }}$

${: : \implies \sf{{- 50 {\bf \: x} + 600= 0 }}}$

${: : \implies \sf{{- 50 {\bf \: x} = - 60 0 }}}$

${: : \implies \sf{{ {\bf \: x} = \frac{ - 600}{ - 50} }}}$

${: : \implies \sf{{ {\bf \: x} = \xcancel\frac{ - 600}{ - 50} }}}$

${: : \implies \sf{{ {\bf \: x} =12cm}}}$

$\bf{Hence,}$

$:: \implies \sf{ QR={ \bf \: x } \: cm = 12 \: cm}$

$\bf{Then,}$

${ : :\implies\sf{PR = 25 \: cm -{ \bf x } } \: \: \: \: \: \: \: \: \: \: \{from \: equation \: (i) \}}$

${ : :\implies\sf{PR = 25 \: cm -{ 12 \: cm} } }$

${ : :\implies\sf{PR =13 \: cm } }$

$\bf{Again,}$

$\sf{The \: value \: of \: sin \: P \: is}$

$\pink{➠} \sf{sin \: P = \frac{side \: opposite \: to \: \angle P}{hypotenuse} }$

$:: \implies\sf{sin \: P = \frac{QR}{PR} }$

$\pink{➠} \sf{sin \: P = \frac{12}{13} }$

$\sf{ And,the\: value \: of \: cos \: P\: is}$

$\pink{➠} \sf{cos \: P = \frac{side \: adjacent \: to \: \angle P}{hypotenuse} }$

$:: \implies\sf{cos \: P = \frac{PQ}{PR} }$

$\pink{➠} \sf{cos \: P = \frac{5}{13} }$

$\sf{At \: last,the\: value \: of \:tan\: P \: is}$

$\pink{➠} \sf{tan \: P = \frac{side \: opposite\: to \: \angle P}{side \: adjacent \: to \: \angle P} }$

$:: \implies\sf{tan \: P = \frac{QR}{PQ} }$

$\pink{➠} \sf{tan \: P = \frac{12}{5} }$

$\sf{If \: we \: use \: alternative \: method \: to \: find \: the \: value \: of \: tan \: P \: is,}$

$\pink{➠} \sf{tan \: P = \frac{sin \: P}{cos \: P} }$

$:: \implies\sf{tan \: P = \frac{ \frac{12}{13} }{ \frac{5}{13} } }$

$:: \implies\sf{tan \: P = \frac{ \frac{12}{ \cancel{13}} }{ \frac{5}{ \cancel{13}} } }$

$\pink{➠} \sf{tan \: P = \frac{12}{5} }$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{green}{Diagram::}}}}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 12cm}\put(2.8,.3){\large\bf 5cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf R}\put(.8,.3){\large\bf Q}\put(5.8,.3){\large\bf P}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

$\large{ \pmb{ \underline{ \underline{\frak{ \color{brown}{Know \: more \: about \: trigonometry:}}}}}}$

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