Question

10. In A 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

PR+QR=25 (given) and PR=25−QR

5

2

+QR

2

=PR

2

(Pythagorean theorem)

Then substitute PR to get this equation = 5

2

+QR

2

=(25−QR)

2

Solve for QR

5

2

+QR

2

=625−50QR+QR

2

(QR

2

cancels out)

5

2

=625−50QR

−600=−50QR

QR=12

Solve for PR using original equation

PR=13

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

sin=

hyp

opp

⇒sin(P)=

13

12

cos=

hyp

adj

⇒cos(P)=

13

5

tan=

adj

opp

⇒tan(P)=

5

12

Answer 2

$\rm { Given \: that, }$

• $\rm { PR + QR = 25}$
• $\rm { PQ = 5 }$

_____

$\rm { Let \: PR \: be \: x.</p><p>}$

$\rm { \therefore QR = 25 - x</p><p> }$

Applying Pythagoras theorem in ΔPQR, we obtain,

$\rm\blue { {PR}^{2} = {PQ}^{2} + {QR}^{2} }$

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

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

$\rm{ \leadsto 50x = 650</p><p> }$

$\rm{ \leadsto x = \cancel {\dfrac{650}{50}</p><p>}=13cm }$

$\rm\red { \therefore, PR = 13 cm</p><p>}$

$\rm\red { QR = (25 - 13) cm = 12 cm }$

______

Now, the values of sin P, cos P and tan P

$\rm { sin \: P =\dfrac{ side \: opposite \: to \: \angle P}{Hypotenuse} = \dfrac{12}{13} }$

$\rm { cos \: P = \dfrac{ side \: adjacent \: to \: \angle P}{Hypotenuse} = \dfrac{5}{13} }$

$\rm { tan \: P= \dfrac{ side \: opposite \: to \: \angle P}{side \: adjacent \: to \: \angle P } = \dfrac{12}{5} }$