Question

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.​

Answer 1

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 !