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 \(\text{tan}\, P.\)

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.

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