Question

In a PQR, right angled
angled at Q, PR+QR = 25
cm and PQ = 5cm. Determine the values
of sin P, cos P and tan P.​

Answer 1

Step-by-step explanation:

sinp=adjacent/hypotenuse

cosp=opp/hypotenuse

tanp=adj/opp

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Answer 2

Given:-

A right angle triangle PQR, right angles at Q.

PR+QR = 25 cm

PQ = 5 cm

Concept:-

• Trigonometry and its applications

Let's Do!

As we are given that PR+QR = 25 cm, we can write it as:-

PR = 25-QR --------------(1)

Now, we will apply Pythagoras Theorem,

$\rm{(Hyp)^2 = (Base)^2+(Height)^2}$

$\rm{(25-QR)^2 = 5^2 + QR^2}$

$\rm{625+QR^2-50QR = 5^2 + QR^2}$

$\rm{625 -50 \ QR = 25}$

$\rm{QR = 12 \ cm}$

Now, we can find RP easily!

$\rm{PR = 25 - 12}$

$\rm{PR = 13 \ cm}$

Now, we can find sinP, cosP and tanP.

We know that:-

$\boxed{\sf{sin \theta = \dfrac{Height}{Hypotenuse}}}$

$\boxed{\sf{cos \theta = \dfrac{Base}{Hypotenuse}}}$

$\boxed{\sf{tan \ \theta = \dfrac{sin \ \theta }{cos \ \theta} \ or \dfrac{Height}{Base} }}$

So, we have :-

Base as QR and height as PQ.

Hypotenuse as PR.

So, sin P = 12/13

And, cos P = 5/13

Tan P = 12/5