Question

In triangle POR, right angle at Q, PR + QR = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tanP.

Answer 1

Given

• ∠Q = 90°
• PR + QR = 25cm
• PQ = 5cm

To Find

• Values of sin P, cos P and tan P.

Solution

Let assume the length of QR = x cm

And PR = (25 - x) cm

Now,

By using Pythagoras Theorem,

RP² = RQ² + QP²

$\tt \implies \: (25 - x) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50x = - 600 \\ \tt \implies \: 50x = 600 \\ \tt \implies x = \frac{600}{50} \\ \tt \implies x = 12$

Then,

QR = x = 12cm

PR = 25 - x = 25 - 12 = 13cm.

∴ sin P = QR/PR = 12/13

cos P = QP/PR = 5/13

and tan P = QR/PQ = 12/5.

Answer 2

${\underline{\huge{\mathbf{\color{pink}{Question}}}}}$

In triangle PQR, right angle at Q, PR + QR = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tanP

${\underline{\huge{\mathbf{\color{pink}{Solution}}}}}$

Given

• ∆PQR is a right angled triangle
• PR + QR = 25cm
• PQ = 5cm

To Find

• sinP
• cosP
• tanP

Step by step solution :-

Using the Pythagorean theorem :-

$\sf{PQ ^{2} + QR {}^{2} =PR {}^{2} }$

$\sf{PQ ^{2} =PR {}^{2} - QR {}^{2} }$

$\sf {(5)}^{2} = (PR + QR)(PR - QR)$

$\sf 25 = 25(PR - QR)$

$\sf PR - QR = 1 \: \: .........(i)$

$\sf{PR + QR = 25}...........(ii)$

Adding (i) and (ii) we have :

$\sf{PR + QR + PR - QR = 25 + 1}$

$\sf2PR = 26$

$\sf{PR = 13}$

Putting the value of PR in (ii) we have :

$\sf 13+ QR = 25$

$\sf{QR = 25 - 13}$

$\sf QR =12$

the sides are :

PQ = 5cm

QR = 12cm

PR = 13cm

Answer :-

$\sf \sin(P) = \frac{QR}{PR}$

• $\sf\sin(P) = \frac{12}{13}$

$\sf\cos(P) = \frac{PQ}{PR}$

• $\sf \cos(P) = \frac{5}{13}$

$\sf\tan(P) = \frac{QR}{PQ}$

• $\sf{ \tan(P) = \frac{12}{5} }$

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