Question

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

Answer 1

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.