Question

In ∆PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Detemine the values of sin P, cos P, tan P ? Class - 10 intro. to trigonometry

Answer 1

• The value of sin P = 12 / 13.
• The value of cos P = 5 / 13.
• The value of tan P = 12 / 5.

Given :

• The base of the right angled triangle = PQ = 5 cm.
• PR + QR = 25 cm.

To Find :

• The value of sin P.
• The value of cos P.
• The value of tan P.

Solution :

We have to find the values of sin P, cos P and tan P.

First, we need to find the measurement of hypotenuse (PR) and perpendicular (QR) of the right angled triangle.

Given,

$\bf \implies PR + QR = 25 cm$

So,

$\bf \implies PR = 25 cm-QR \: \: ......(1)$

By the pythagoras theorem,

$\bf \implies {P}^{2} + {B}^{2} = {H}^{2} \\ \\ \\ \bf \implies {QR}^{2} + {PQ}^{2} = {PR}^{2}$

Where,

• H = Hypotenuse = PR = 25 - QR.
• P = Perpendicular = QR.
• B = Base = PQ = 5 cm.

Now, substitute all the values in the pythagoras theorem.

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

⠀

By using identity,

$\huge \star \: \: \: \small\bf \bigg( \: {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \bigg)$

⠀

$\bf \implies {QR}^{2} + 25 \: {cm}^{2} = {(25 \: cm)}^{2} - 2(25 \: cm)(QR) + {(QR)}^{2} \\ \\ \\ \bf \implies \not{QR}^{2} + 25\: {cm}^{2} = 625 \: {cm}^{2} - 50 \: QR \: cm \: \: + \not{QR}^{2} \\ \\ \\ \bf \implies 25 \: {cm}^{2} = 625 \: {cm}^{2} - 50 \: QR \: cm\\ \\ \\ \bf \implies 50 \: QR \: cm = 625 \: {cm}^{2} - 25 \: {cm}^{2}\\ \\ \\ \bf \implies 50 \: QR \: cm = 600 \: {cm}^{2} \\ \\ \\ \bf \implies QR \: = \dfrac{\not600 \: {cm}^{2} }{\not50 \: cm} \\ \\ \\\bf \implies QR = 12 \: cm \\ \\ \\ \: \: \therefore \bf \: \: Perpendicular= 12 \: cm$

Now, substitute the value of QR in the equation (1),

$\bf \implies PR = 25 cm-QR \\ \\ \\ \bf \implies PR = 25 \:cm-12 \: cm \\ \\ \\ \bf \implies PR = 13 \: cm \\ \\ \\ \: \: \bf \therefore \: \: Hypotenuse = 13 \: cm$

Now, we have to find the values of sin P, cos P and tan P.

1) sin P.

We know that,

$\huge \blue{\star} \: \: \small \bf sin \: \theta = \dfrac{Perpendicular}{Hypotenuse}$

So,

$\bf \implies sin \: p = \dfrac{QR}{PR} \\ \\ \\ \bf \implies sin \: p = \dfrac{12}{13}$

2) cos P.

We know that,

$\huge \blue{\star} \: \: \small \bf cos \: \theta = \dfrac{Base}{Hypotenuse}$

So,

$\bf \implies cos \: p = \dfrac{PQ}{PR} \\ \\ \\ \bf \implies cos \: p = \dfrac{5}{13}$

3) tan P.

We know that,

$\huge \blue{\star} \small \bf \: \: \: tan \: \theta = \dfrac{Perpendicular}{Base}$

So,

$\bf \implies tan \: p = \dfrac{QR}{PQ} \\ \\ \\ \bf \implies tan \: p = \dfrac{12}{5}$

Hence,

The value of sin P is 12 / 13, cos P is 5 / 13 and tan P is 12 / 5.

Answer 2

Answer:

Given :-

• In ∆ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm.

To Find :-

• What is the value of sin P, cos P, tan P.

Solution :-

In ∆ PQR,

➟ PR = QR = 25 cm

➟ PR = (25 + QR)

And, PQ = 5 cm

By using Phythagorus Theorem,

$\boxed{\bold{\small{Q{R}^{2} = P{Q}^{2} + P{R}^{2}}}}$

⇒ QR² = (5)² + (25 + QR)²

⇒ QR² = 25 + 625 + 50 QR + QR

⇒ 25 + QR = 625 - 50 QR + QR

⇒ 50 + QR = 625 - 25

⇒ 50 + QR = 600

⇒QR = $\sf\dfrac{\cancel{600}}{\cancel{50}}$

➟ QR = 12 cm

Hence,

• QR (Perpendicular) = 12 cm
• PR (Hypotenuse) = (25 - QR) = (25 - 12) = 13 cm
• PQ (Base) = 5 cm (Given)

Now, we have to find the values of sin P, cos P, tan P,

❶ sin P = $\dfrac{p}{h}$ = $\dfrac{QR}{PR}$ = $\boxed{\bold{\small{\dfrac{12}{13}}}}$

❷ cos P = $\dfrac{b}{h}$ = $\dfrac{PQ}{PR}$ = $\boxed{\bold{\small{\dfrac{5}{13}}}}$

❸ tan P = $\dfrac{p}{b}$ = $\dfrac{QR}{PQ}$ = $\boxed{\bold{\small{\dfrac{12}{5}}}}$

$\therefore$ The value of sin P is $\dfrac{12}{13}$, cos P is $\dfrac{5}{13}$ and tan P is $\dfrac{12}{5}$.