Question

​$\boxed {\boxed{ { \red{ \bold{ \underline{Question:-}}}}}}$

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

Answer 1

$\huge{\underline{\mathcal{\red{R}\pink{e}\green{q}\blue{u}\purple{i}\orange{r}\purple{e}\pink{d-}\red{A}\blue{n}\pink{s}\green{w}\purple{e}\orange{r}}}}$

Given:-

PR + QR = 25, PQ = 5

PR be x. and QR = 25 -x

Pythagoras theorem, a

PR² = PQ² + QR²

x² = (5) ²+(25-x) ²

x² = 25 + 625 + x² - 50x

50x = 650

x = 13

PR = 13 cm

QR = (25 - 13) cm = 12cm

sin P = QR/PR = 12/13

sin P = PQ/PR = 5/13

sin P = QR/PQ = 12/15

❤Hope It Helps You ❤

Answer 2

$\large \sf \underline \bold{ \underline {Diagram : }}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{p}}\put(7.7,1){\large{Q}}\put(10.6,1){\large{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\sf{\large{5 cm}}}\put(9,0.7){\sf{\large{x cm}}}\put(9.4,1.9){\sf{\large{(25-x) cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\large \sf \underline \bold{ \underline {Solution: }}$

$\rm\:let\:QR=x\:cm$

Now,

$\rm\:PR+QR=25$

$\rm\implies\:PR+x=25$

$\rm\implies\:PR=25-x$

Then,

$\rm\:In\:right{\triangle\:PQR}$

$\rm\:{(PR)}^{2}={(PQ)}^{2}+{(QR)}^{2}$

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

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

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

$\rm\implies\:625-50x=25+\cancel{x}^{2}-\cancel{x}^{2}$

$\rm\implies\:625-50x=25$

$\rm\implies\:625-25=50x$

$\rm\implies\:600=50x$

$\rm\implies\:x=\cancel\dfrac{600}{50}$

$\rm\implies\:x=12$

So,

$\rm\:QR=x=12\:cm$

$\rm\:PR=25-x$

$\rm\:PR=25-12$

$\rm\:PR=13\:cm$

According to the question:-

$\rm\:sinP=\dfrac{p}{h}=\dfrac{QR}{PR}$

$\rm\:sinP=\dfrac{12}{13}$

$\rm\:cosP=\dfrac{b}{h}=\dfrac{PQ}{PR}$

$\rm\:cosP=\dfrac{5}{13}$

$\rm\:tanP=\dfrac{p}{b}=\dfrac{QR}{PQ}$

$\rm\:tanP=\dfrac{12}{5}$

So,

Final Diagram:-

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{p}}\put(7.7,1){\large{Q}}\put(10.6,1){\large{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\sf{\large{5 cm}}}\put(9,0.7){\sf{\large{12 cm}}}\put(9.4,1.9){\sf{\large{13 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$