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# In $\Delta$PQR, Right-angled at Q. PR+QR=$25$cm and PQ=$5$cm. Determine the values of ${\text{sin P, cos P, tan P}}{\text{.}}$

Last updated date: 12th Sep 2024
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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.