Question

Answer:

Step-by-step explanation:

Answer:-

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}(Hyp)2=(Base)2+(Height)2

\rm{(25-QR)^2 = 5^2 + QR^2}(25−QR)2=52+QR2

\rm{625+QR^2-50QR = 5^2 + QR^2}625+QR2−50QR=52+QR2

\rm{625 -50 \ QR = 25}625−50 QR=25

\rm{QR = 12 \ cm}QR=12 cm

Now, we can find RP easily!

\rm{PR = 25 - 12}PR=25−12

\rm{PR = 13 \ cm}PR=13 cm

Now, we can find sinP, cosP and tanP.

We know that:-

\boxed{\sf{sin \theta = \dfrac{Height}{Hypotenuse}}}sinθ=HypotenuseHeight​​

\boxed{\sf{cos \theta = \dfrac{Base}{Hypotenuse}}}cosθ=HypotenuseBase​​

\boxed{\sf{tan \ \theta = \dfrac{sin \ \theta }{cos \ \theta} \ or \dfrac{Height}{Base} }}tan θ=cos θsin θ​ orBaseHeight​​

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

And hence, are the answers.​

Answer 1

Answer:

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