Question

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

NCERT Class X
Mathematics - Mathematics

Chapter _INTRODUCTION TO
TRIGONOMETRY

Answer 1

DIAGRAM OF THE QUESTION ARE GIVEN BELOW IN ATTACHMENT

It is given that PQR is a right Δ, such that <Q = 90 °
                            PR +QR = 25 cm
         and              PQ = 5 cm
         Let,              QR = x cm
         therefore,      PR = (25-x)
         therefore,      By Pythagoras Theorem, we have
                            PR² = QR² + PQ²
         ⇒                (25-x)² = x² + 5²
         ⇒                625 - 50x + x² = x² + 25
         ⇒                -50x = -600
         ⇒                 x = -600/-50 = 12
         i.e.,               PQ = 12 cm
         ⇒                  RP = 25 - 12 = 13 cm
         Now, we have:
                               sin P = RQ/RP = 12/13
                               cos P = PQ/RP = 5/13
                               tan P = RQ/PQ = 12/5
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Answer 2

$\rm { Given \: that, }$

$\rm { PR + QR = 25}$

$\rm { PQ = 5 }$

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

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

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

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

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

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