Question

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

Answer 1

Step-by-step explanation:

First you have to solve for QR and PR.

PR+QR = 25 (given) and PR=25-QR

52 + QR2= PR2 (pythagorean theorem)

Then substitute PR to get this equation = 52 + QR2 = (25-QR)2

Solve for QR

52 + QR2 = 625 - 50QR +QR2 .........(QR2 cancels out)

52 = 625 - 50 QR

-600 = -50 QR

QR= 12

Solve for PR using original equation.

PR= 13

Now draw the triangle on your paper for help solving the next step...

Sin = opposite/ hypotenuse Sin(P)= 12/13

Cos= Adjacent/ hypotenuse Cos(P) = 5/13

Tan= opposite/ adjacent Tan(P) = 12/5

Answer 2

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

$\huge{\mathfrak{\green{\orange{\bf\:S}\pink{o}\purple{l}\green{u}\orange{t}\blue{i}\blue{o}\red{n:-}}}}$

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

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:-

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