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Question: In \[\Delta {\text{PQR}}\] , right-angled at \({\text{Q}}\) , \({\text{PR}} + {\text{QR}} = 25{\text...

In ΔPQR\Delta {\text{PQR}} , right-angled at Q{\text{Q}} , PR+QR=25cm{\text{PR}} + {\text{QR}} = 25{\text{cm}} and PQ=5cm{\text{PQ}} = {\text{5cm}}. Determine the values of sinP\sin {\text{P}} , cosP\cos {\text{P}} and tanP\tan {\text{P}}.

Explanation

Solution

In this question first we will draw a triangle ΔPQR\Delta {\text{PQR}} having Q{\text{Q}} as a right angle. Now we will find the length of QR{\text{QR}} with the help of the equation given in the question and Pythagoras theorem. Now, with the help of QR{\text{QR}} and PR{\text{PR}}. We will find what is asked in the question.

Complete step-by-step solution:
We know that PQ=5cm{\text{PQ}} = {\text{5}}\,cm (given in the question). Now we have to find the remaining sides of the triangle ΔPQR\Delta {\text{PQR}}

Now, we know that PR+QR=25{\text{PR}} + {\text{QR}} = 25 (given in the question). Therefore, it can be written as PR=25QR{\text{PR}} = 25 - {\text{QR}}. From PR=25QR{\text{PR}} = 25 - {\text{QR}}we will find the length of QR{\text{QR}}.
From the figure we can write {\text{P}}{{\text{R}}^{\text{2}}} = {\text{Q}}{{\text{R}}^{\text{2}}} + {\text{P}}{{\text{Q}}^{\text{2}}}\\_\\_\\_(1)
Put the values of PQ{\text{PQ}} and PR{\text{PR}} in equation (1)
(25QR)2=QR2+(5)2\Rightarrow {\left( {{\text{25}} - {\text{QR}}} \right)^{\text{2}}} = {\text{Q}}{{\text{R}}^{\text{2}}} + {\left( {\text{5}} \right)^{\text{2}}}
Now, simplify the above equation
62550QR+QR2=QR2+25\Rightarrow {\text{625}} - {\text{50QR}} + {\text{Q}}{{\text{R}}^{\text{2}}} = {\text{Q}}{{\text{R}}^{\text{2}}} + {\text{25}}
Now, cancel the common things
62550QR25=0\Rightarrow {\text{625}} - {\text{50QR}} - {\text{25}} = {\text{0}}
Now, we can easily find QR{\text{QR}}
50QR=600 QR=12 \Rightarrow 50{\text{QR}} = 600 \\\ \Rightarrow {\text{QR}} = 12
Therefore, we found the length of QR{\text{QR}} is 12cm12{\text{cm}} . Now, we know that PR=25QR{\text{PR}} = 25 - {\text{QR}}.
PR=25QR PR=2512=13 {\text{PR}} = 25 - {\text{QR}} \\\ \Rightarrow {\text{PR}} = 25 - 12 = 13
From the above calculation we found the length of PR{\text{PR}} is 13cm13{\text{cm}}.
Now, we know that sinP=opposite sidehypotenuse\sin {\text{P}} = \dfrac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} . From the figure we can write sinP=QRPR\sin {\text{P}} = \dfrac{{{\text{QR}}}}{{{\text{PR}}}}.
Therefore, we can write sinP=QRPR=1213\sin {\text{P}} = \dfrac{{{\text{QR}}}}{{{\text{PR}}}} = \dfrac{{12}}{{13}}.
Similarly, we know that cos P=Adjacent sidehypotenuse{\text{cos P}} = \dfrac{{{\text{Adjacent side}}}}{{{\text{hypotenuse}}}} . From the figure we can write cos P=PQPR{\text{cos P}} = \dfrac{{{\text{PQ}}}}{{{\text{PR}}}}.
Therefore, we can write cos P=PQPR=513{\text{cos P}} = \dfrac{{{\text{PQ}}}}{{{\text{PR}}}} = \dfrac{5}{{13}}.
Similarly, we know that tan P=opposite sideAdjacent side{\text{tan P}} = \dfrac{{{\text{opposite side}}}}{{{\text{Adjacent side}}}} . From the figure we can write tan P=QRPQ{\text{tan P}} = \dfrac{{{\text{QR}}}}{{{\text{PQ}}}}.
Therefore, we can write tan P=QRPQ=125{\text{tan P}} = \dfrac{{{\text{QR}}}}{{{\text{PQ}}}} = \dfrac{{12}}{5}.

Hence, the answers are sinP=1213\sin {\text{P}} = \dfrac{{12}}{{13}}, cos P=513{\text{cos P}} = \dfrac{5}{{13}} and tan P=125{\text{tan P}} = \dfrac{{12}}{5}.

Note: We can also solve this question by making QR{\text{QR}} as a subject and then from it we can find PR{\text{PR}} first. The other important things are the formula of sin\sin , cos\cos and tan\tan which we need to memorize.
sinA = OppositeHypotenuse\sin {\text{A = }}\dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}
cosA = AdjacentHypotenuse\cos {\text{A = }}\dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}
tanA = OppositeAdjacent\tan {\text{A = }}\dfrac{{{\text{Opposite}}}}{{{\text{Adjacent}}}}
cosecA = HypotenuseOpposite\cos ec {\text{A = }}\dfrac{{{\text{Hypotenuse}}}}{{{\text{Opposite}}}}
secA = HypotenuseAdjacent\sec {\text{A = }}\dfrac{{{\text{Hypotenuse}}}}{{{\text{Adjacent}}}}
cotA = AdjacentOpposite\cot {\text{A = }}\dfrac{{{\text{Adjacent}}}}{{{\text{Opposite}}}}