Question

9. In ∆PQR right angled at Q, PQ = 5 cm and <R = 30°, find QR and PR.
the correct ans is QR =5√3
PR= 10cm​

Answer 1

Question : -

In ∆ PQR , right angled at Q . PQ = 5 cm and ∠R = 30° , find the measurement of side QR & PR ?

Answer : -

Given : -

In ∆ PQR , right angled at Q . PQ = 5 cm and ∠R = 30° .

Required to find : -

• Measurement of sides QR & PR ?

Trigonometric ratio used : -

$\boxed{\tt{\bf{ \tan \theta = \dfrac{Opposite \ side}{Adjacent \ side} }}}$

$\boxed{\tt{\bf{ \sin \theta = \dfrac{Opposite \ side}{ Hypotenuse} }}}$

Solution : -

- : Diagram : -

$\setlength{ \unitlength}{20} \begin{picture}(6,6) \linethickness{1} \put(1,1){\line(0,1){4}}\put(1,1){\line(1,0){3}}\qbezier(4,1)(4,1)(1,5)\put(1.25,1){\line(0,1){0.3}}\put(1,1.3){\line(1,0){0.25}}\put(4,0.5){ \bf P }\put(0.8,0.5){ \bf Q }\put(1,5.25){ \bf R }\put(2,0.5){ \sf 5 \: cm }\qbezier(1,4.5)(1.25,4)(1.4,4.5)\put(1.1,3.5){ \bf {30}^{ \circ} }\end{picture}$

In ∆ PQR , right angled at Q . PQ = 5 cm and ∠R = 30° .

we need to find the measurement of sides QR & PR .

So,

Here,

From the diagram we can conclude that ;

Using tan 30° we can find the value of side QR .

Since,

we know that ;

$\boxed{\tt{\bf{ \tan \theta = \dfrac{Opposite \ side}{Adjacent \ side} }}}$

Here,

$\theta$ = 30°

This implies ;

$\boxed{\tt{\bf{ \tan {30}^{\circ} = \dfrac{Opposite \ side}{Adjacent \ side} }}}$

• Opposite side = QP

• Adjacent side = PQ

➔ Tan 30° = QP/PQ

➔ ( As QP = 5 cm )

➔ Tan 30° = 5 / PQ

➔ From the standard Completementary angles we can say that ;

➔ Tan 30° = 1/√3

So,

➔ 1/√3 = 5/PQ

cross multiplication

➔ PQ = 5√3

Hence,

• Measurement of side PQ = 5√3 cm or 8.66 cm ( approximately )

Similarly,

We know that ;

$\boxed{\tt{\bf{ \sin \theta = \dfrac{Opposite \ side}{ Hypotenuse} }}}$

This implies ;

➔ Sin 30° = Opposite side/ Hypotenuse

( $\theta$ = 30° )

Here,

• Opposite side = QP

• Hypotenuse = PR

➔ Sin 30° = QP/PR

( since, QP = 5cm )

➔ Sin 30° = 5/PR

From the standard Completementary angles of trigonometry we can say that ;

➔ Sin 30° = 1/2

➔ 1/2 = 5/PR

➔ PR = 10 cm

Hence,

• Measurement of side PR = 10 cm

Answer 2

◘ Given ◘

A right angled triangle PQR with :

• ∠Q = 90°
• PQ = 5 cm
• ∠R = 30°

◘ To Find ◘

The length of QR and PR.

◘ Solution ◘

$\bigstar \underline{\underline{\sf{DIAGRAM:-}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{P}}\put(7.7,1){\large{Q}}\put(10.6,1){\large{R}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(7.3,2){\sf{\large{5\ cm}}}\put(9,0.7){\sf{\large{}}}\put(9.4,1.9){\sf{\large{}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

We know,

► sinθ = Perpendicular / Hypotenuse

→ sin R = PQ / PR

→ sin 30° = 5 / PR

→ 1/2 = 5 / PR

→ 2 = PR/5

→ PR = 10 cm

_____________________

Applying Pythagoras Theorem,

$\sf{PR^2=PQ^2+QR^2}\\\\\sf{\longrightarrow (10)^2=(5)^2+QR^2}\\\\\sf{\longrightarrow 100=25+QR^2}\\\\\sf{\longrightarrow QR^2=100-25}\\\\\sf{\longrightarrow QR^2=75}\\\\\sf{\longrightarrow QR=\sqrt{75}}\\\\\sf{\longrightarrow QR=\sqrt{25\times3} }\\\\\sf{\longrightarrow QR=5\sqrt{3}\ cm}$

_____________________

Therefore,

• PR = 10 cm

• QR = 5√3 cm