Question

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

Answer 1

Given that, PR + QR = 25
PQ = 5

Let PR be x

Therefore,

QR = 25 - x



Applying Pythagoras theorem in ΔPQR, we obtain

PR2 = PQ2 + QR2

x2 = (5)2 + (25 - x)2

x2 = 25 + 625 + x2 - 50x          [as, (a + b)2 = a2+ b2 + 2ab]

50x = 650

x = 13

Therefore,

PR = 13 cm

QR = (25 - 13) cm = 12 cm
and we know, 

Sin P = 

Cos P = 

Tan P = 

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

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

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

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