Question

Q). 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.
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Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

In ∆PQR , right angled at Q

• PR+QR=25cm
• PQ=5cm

${\bf{\blue{\underline{Prove:}}}}$

• SinP,CosP and TanP

${\bf{\blue{\underline{Now:}}}}$

$\star \: {\bf{PR + QR = 25cm}}$

$\star \: {\bf{PR = 25-QR}}$

In right angled ∆PQR,

$\implies\: {\bf{ {PR}^{2} = {PQ}^{2} + {QR}^{2} }}$

$\implies\: {\bf{ {(25 - QR)}^{2} = {(5)}^{2} + {(QR)}^{2} }}$

$\implies\: {\bf{ ( {25)}^{2} - 2(25)QR + ( {QR)}^{2} = 25 + {(QR)}^{2} }}$

$\implies\: {\bf{ 625 - 50QR + ( {QR)}^{2} = 25 + {(QR)}^{2} }}$

$\implies\: {\bf{ 625 - 50QR + ( {QR)}^{2} - ( {QR)}^{2} - 25 }}$

$\implies\: {\bf{ 625 - 50QR - 25 = 0 }}$

$\implies\: {\bf{ 600 - 50QR = 0 }}$

$\implies\: {\bf{ - 50QR = - 600}}$

$\implies\: {\bf{ QR = \frac{600}{50} }}$

$\implies{ \boxed{ \purple {\bf{ QR = 12}}}}$

Now,

• PR =25-QR
• PR =25-12
• PR = 13

$\star{\boxed{\bf{ SinP = \frac{P}{H} = \frac{5}{13} }}}$

$\star{\boxed{\bf{ CosP = \frac{B}{H} = \frac{12}{13} }}}$

$\star{\boxed{\bf{ TanP= \frac{P}{B} = \frac{5}{12} }}}$

Answer 2

Answer:

Sin P=12/13  , tan P=12/5

Step-by-step explanation:

Let PR be 'x' and QR be = 25-x

Using Pythagorus Theorem,

==>PR^2 = PQ^2 + QR^2

     x^2   = (5)^2 + (25-x)^2

     x^2   = 25 + 625 + x^2 - 50x

     50x   = 650

          x   = 13

therefore, PR = 13 cm

and,          QR = (25 - 13) = 12 cm

Now,  Sin P = opposite/hypotenuse = QR/PR = 12/13

          Tan P= opposite/adjacent      = QR/PQ = 12/5

          Cos P= adjacent/hypotenuse = PQ/PR = 5/13