Math, asked by SuhaniJaiswar, 1 month ago

In Δ PQR, right-angled at Q, PR + QR = 30 cm and PQ = 10 cm. Determine the values of sin P, cos P and tan P.​

Answers

Answered by Yuseong
17

Formulae to use :

 \sf { \sin P = \dfrac{Perpendicular}{Hypotenuse} }

 \sf { \cos P = \dfrac{Base}{Hypotenuse} }

 \sf { \tan P = \dfrac{Perpendicular}{Base} }

Pythagoras property :

  • H² = B² + P²

● H = Hypotenuse

● B = Base

● P = Perpendicular

Algebraic identity :

  • (a - b)² = a² + b² - 2ab

Explication of steps :

Given that :

  • PQR is a ∆, right-angled at Q.
  • PR + QR = 30 cm
  • PQ = 10 cm

To calculate :

  • The values of sin P, cos P and tan P.

Calculation :

According to the question,

\longmapsto \: \boxed{\bf {Base(PQ) =10 \: cm }} \\

\longrightarrow \sf { PR + QR = 30 \: cm }

Let, PR be x cm. So,

\longrightarrow \sf { x + QR = 30  \: cm}

\longrightarrow \sf { QR =( 30 -x ) cm }

• PR = Hypotenuse (x cm)

• PQ = Base (10 cm)

• QR = Perpendicular → ( 30 - x ) cm

 \underline{\small \sf {\maltese \; \; \; Finding \: value \:  of \: PR \: and \: QR : \; \; \;  }}

By using pythagoras property,

 \longrightarrow H² = B² + P²

 \longrightarrow (x)² = (10)² + (30 - x)²

  • (a - b)² = a² + b² - 2ab

 \longrightarrow (x)² = (10)² + (30)² + (x)² - 2(30x)

 \longrightarrow x² = 100 + 900 + x² - 60x

 \longrightarrow x² - x² = 1000 - 60x

 \longrightarrow 0 = 1000 - 60x

 \longrightarrow 0 + 60x = 1000

 \longrightarrow 60x = 1000

 \longrightarrow x =  \sf \dfrac{ 1000}{60}

 \longrightarrow x =  \sf \dfrac{ 100}{6}

 \longrightarrow x =  \sf \dfrac{ 50}{3}

So,

\longmapsto \: \boxed{\bf {Hypotenuse(PR) = \dfrac{50}{3} \: cm }} \\

Also,

 \longrightarrow QR = ( 30 - x ) cm

 \longrightarrow QR = ( 30 -  \sf \dfrac{ 50}{3} ) cm

 \longrightarrow QR =  \sf \dfrac{ 90 - 50}{3} cm

 \longrightarrow QR =  \sf \dfrac{ 40}{3} cm

\longmapsto \: \boxed{\bf {Perpendicular(QR) = \dfrac{40}{3} \: cm }} \\

Value of sin P , cos P and tan P :

 \sf { \sin P = \dfrac{Perpendicular}{Hypotenuse} }

 \longmapsto \sf { \sin P = \dfrac{QR}{PR} }

 \longmapsto \sf {\sin P = \dfrac{\cfrac{40}{3}}{\cfrac{50}{3} }  }

 \longmapsto \sf {\sin P = \dfrac{40}{3}\times \dfrac{3}{50} }

 \longmapsto \sf {\sin P = \dfrac{40}{50} }

 \longmapsto \underline { \boxed { \bf {\sin P = \dfrac{4}{5} }}}

_________

 \sf { \cos P = \dfrac{Base}{Hypotenuse} }

 \longmapsto \sf { \cos P = \dfrac{PQ}{PR} }

 \longmapsto \sf {\cos P = \dfrac{10}{\cfrac{50}{3} }  }

 \longmapsto \sf {\cos P = 10 \times \dfrac{3}{50} }

 \longmapsto \sf {\cos P = 1\times \dfrac{3}{5} }

 \longmapsto \underline { \boxed { \bf {\cos P = \dfrac{3}{5} }}}

_________

 \sf { \tan P = \dfrac{Perpendicular}{Base} }

 \longmapsto \sf { \tan P = \dfrac{QR}{PQ} }

 \longmapsto \sf {\sin P = \dfrac{\cfrac{40}{3}}{10 }  }

 \longmapsto \sf {\tan P = \dfrac{40}{3}\times \dfrac{1}{10} }

 \longmapsto \sf {\tan P = \dfrac{4}{3}\times \dfrac{1}{1} }

 \longmapsto \underline { \boxed { \bf {\tan P = \dfrac{4}{3} }}}

Hence, we got the answer !

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