Question

In ∆PQR, right angled at Q, PR + QR = 25cm and PQ = 5cm. Determine the values of -
(i) sinP
(ii) cosP and
(iii) tanP​

Answer 1

Answer:

i) 12/13

ii) 5/13

iii) 12/5

Step-by-step explanation:

Let QR be x

PR + x = 25

PR = 25 - x

By Pythagoras Theorem,

Square of Hypotenuse = Sum of squares of other two sides

PR² = PQ² + QR²

(25 - x)² = x² + 5²

25² - 50x + x² = x² + 25

625 - 50x = 25

- 50x = 25 - 625

- 50x = - 600

50x = 600

x = 600/50

x = 12 cm

QR = 12 cm

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

i) sinP = opposite/hypotenuse

= 12/13

ii) cosP = adjacent/hypotenuse

= 5/13

iii) tanP = opposite/adjacent

= 12/5

Hope it helps you:)

Answer 2

1. $\fbox \red { sin(p) = 5/13 }$
2. $\fbox \purple { cos(p) = 12/13 }$
3. $\fbox \green { tan(p) = 5/12 }$

Step-by-step explanation:

Hi ! Here u go with your Answer

GIVEN :-

PQ = 5cm

Let QR be x and

PR= ( 25-x)

APPLYING PYTHAGORAS THEOREM:-

Hypotenuse ² =v Perpendicular ² + Base²

[let's put the values]

${(25 - x)}^{2} = {5}^{2} + {x}^{2} \\ \\ \\ 625 + {x}^{2} - 50x = {x}^{2} + 25 \\ \\ 625 - 50x = 25 \\ \\ - 50x = 25 - 625 \\ \\ - 50x = - 600 \\ \\ x = \frac{ - 600}{ - 50} \\ \\ x = 12 \: cm$

Hence the value of x is 12 cm

let's substitute the values

PR = 25 -x

= 25 - 12

= 13 cm

And QR = 12cm

____________________________

$\sin(p) = \frac{perpendicular}{hypotenuse} \\ \\ = \frac{5}{13}$

________________________

$\cos(p) = \frac{base}{hypotenuse} \\ \\ = \frac{12}{13}$

_______________________

$\tan(p) = \frac{ \sin(p) }{ \cos(p) } \\ \\ = \frac{ \frac{5}{13} }{ \frac{12}{13} } \\ \\ = \frac{5 \times 13}{12 \times 13 } \\ \\ = \frac{5}{12}$