Answer

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**Hint:**In this type of question, first we need to assume a variable for unknown.

Then we can easily solve them by using the Pythagoras theorem.

**Complete step-by-step answer:**

It is given that PR+ QR= \[25\]cm,

PQ= \[5\]cm

In this question we need to find the values of \[{\text{sin P, cos P}}\] and\[{\text{tan P}}\].

Let us assume that PR be \[x\]

We Know that,

PR+ QR=\[25\]cm

Substituting \[x\] for PR

\[ \Rightarrow \] \[x + \] QR \[ = 25\]

From the given data we can written as,

QR \[ = 25 - x\]

Now we know that

PR \[{\text{ = }}\] \[x\]

QR \[ = 25 - x\]

PQ \[{\text{ = }}\] \[{\text{5}}\]

In Right-angled triangle PQR,

Here we using the Pythagoras theorem,

\[P{R^2} = P{Q^2} + Q{R^2}\]

Here, we are substituting the above values,

\[{x^2} = {5^2} + {\left( {25 - x} \right)^2}...\left( 1 \right)\]

Now, we are expanding the above expansion using the formula. i.e. \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]

Here \[a = 25\]and \[{\text{ }}b = x\;\]

\[{\left( {25 - x} \right)^2} = 625 + {x^2} - 50x\]

Now we can write it as, and squaring first term in the LHS in \[\left( 1 \right)\] ,we get

\[{x^2} = 25 + 625 + {x^2} - 50x\]

On cancelling the same term and add the two terms as RHS we get that,

\[50x = 650\]

On dividing \[50\] on both side we get,,

\[x = \dfrac{{650}}{{50}}\]

On dividing we get,

\[x = 13\]

\[\therefore PR = 13cm,\]

Now we got the value of PR that is the value of the \[x\].

Also we are going to substitute the value that we got,

\[QR = 25 - x\]

\[QR = 25 - 13\]

By subtracting we have that,

\[QR = 12cm\]

Here According to the question,

We need to find \[{\text{sin P, cos P}}\]and\[{\text{tan P}}\].

For that we are going to substitute the values we get on the Formulas.

Sin P \[ = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}\] \[ = \dfrac{{QR}}{{PR}} = \dfrac{{12}}{{13}}\]

Cos P \[ = \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}\] \[ = \dfrac{{PQ}}{{PR}} = \dfrac{5}{{13}}\]

tan P \[ = \dfrac{{{\text{Opposite}}}}{{{\text{Adjacent}}}}\] \[ = \dfrac{{QR}}{{PQ}} = \dfrac{{12}}{5}\]

Finally, here the answers we got are,

Sin P \[ = \dfrac{{12}}{{13}}\]

Cos P \[ = \dfrac{5}{{13}}\]

tan P \[ = \dfrac{{12}}{5}\]

**Note:**According to Pythagoras theorem, the square of the hypotenuse side is equal to the sum of the squares of the other two sides in a right-angled triangle.

Mostly In this type of questions like Right-Angled triangle will be interrelated with Pythagoras theorem. You can solve them easily using this theorem.

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