# Ex.8.1 Q10 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

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

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 10

## Text Solution

#### Reasoning:

Using Pythagoras theorem, we can find the length of the all three sides. Then the required trignometric ratios

#### Steps:

Given, $$\text{ }\!\!\Delta ABC$$ be a right angle at $${Q.}$$

\begin{align} & {PQ=5}\,\text{cm} \\ & {PR+QR=25} \\ \end{align}

Let $${PR}= x \, \text{cm}$$

Therefore,

\begin{align} QR\, &=25\,\text{cm}- PR \\ &=25\,\text{cm}-x\,\text{cm} \end{align}

By applying Pythagoras theorem for \begin{align}\Delta {PQR}\end{align} we obtain.

\begin{align} P{{R}^{2}}&=P{{Q}^{2}}+Q{{R}^{2}} \\ {{x}^{2}}&={{(5)}^{2}}+{{(25-x)}^{2}} \\ {{x}^{2}}&=25+625-50x+{{x}^{2}} \\ 50x&=650 \\ x&=\frac{650}{50} \\ &=130\ \rm{ m} \end{align}

Therefore,

\begin{align} {PR} &=13 \rm{cm} \\ {QR} &=(25-13) \rm{cm} \\ &=12 \rm{cm} \end{align}

By substituting the values obtained above in the trigonometric functions below.

\begin{align}\sin {P}&=\frac{\text { side opposite to } \angle {P}}{\text { hypotenuse }}\\ &=\frac{{QR}}{{PR}}=\frac{12}{13} \\\\ \cos {P} & =\frac{\text { side adjacent to } \angle {P}}{\text { hypotenuse }} \\ & =\frac{{PQ}}{{PR}}=\frac{5}{13} \\\\ \tan {P} & =\frac{\text { side opposite to } \angle {P}}{\text { side adjacent to } \angle {P}} \\ & =\frac{{QR}}{{PQ}}=\frac{12}{5}\end{align}

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