Question

In the figure of $\Delta PQR,\angle P=\theta {}^\circ \text{ and }\angle R=\phi {}^\circ $ . Find
(i) $\sqrt{x+1}\cot \phi $
(ii) $\sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta $
(iii) $\cos \theta $

Answer 1

Hint: We have to apply the Pythagoras theorem on the given right-angled triangle which states that the square of the hypotenuse of a right-angled triangle is equal to the sum of squares of the other two sides and find the value of PQ. Now, to find the value of $\sqrt{x+1}\cot \phi $ , we have to use the formula $\cot \phi =\dfrac{\text{Adjacent side}}{\text{Opposite side}}$ and simplify the expression. Similarly, for $\sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta $ , we have to substitute $\tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}$ and simplify. Finally, $\cos \theta =\dfrac{\text{Adjacent side}}{\text{Hypotenuse}}$ gives the value of (iii).

Complete step by step solution:
We are given with a triangle $\Delta PQR$ with $\angle P=\theta {}^\circ \text{ and }\angle R=\phi {}^\circ $ . From the given figure, we can see that $\angle Q=90{}^\circ $ , that is, the $\Delta PQR$ is a right-angled triangle. We can now apply the Pythagoras theorem which states that the square of the hypotenuse of a right-angled triangle is equal to the sum of squares of the other two sides. In the given figure, we can see that PR is the hypotenuse. Therefore, we can write
$\Rightarrow P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}$
From the given figure, we can see that $PR=x+2$ and $QR=x$ .
$\Rightarrow {{\left( x+2 \right)}^{2}}=P{{Q}^{2}}+{{x}^{2}}$
Let us find PQ. We have to take ${{x}^{2}}$ to the LHS.
$\begin{align}
  & \Rightarrow {{\left( x+2 \right)}^{2}}-{{x}^{2}}=P{{Q}^{2}} \\
 & \Rightarrow P{{Q}^{2}}={{\left( x+2 \right)}^{2}}-{{x}^{2}} \\
\end{align}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, the above equation can be written as
$\begin{align}
  & \Rightarrow P{{Q}^{2}}={{x}^{2}}+2\times 2x+4-{{x}^{2}} \\
&\Rightarrow P{{Q}^{2}}=\require{cancel}\cancel{{{x}^{2}}}+4x+4\require{cancel}\cancel{-{{x}^{2}}} \\
 & \Rightarrow P{{Q}^{2}}=4x+4 \\
\end{align}$
Let us take 4 outside from the terms in the RHS.
$\Rightarrow P{{Q}^{2}}=4\left( x+1 \right)$
We have to take square roots on both sides.
$\begin{align}
  & \Rightarrow PQ=\sqrt{4\left( x+1 \right)} \\
 & \Rightarrow PQ=2\sqrt{\left( x+1 \right)}...\left( i \right) \\
\end{align}$
Now, let us find the values of each of the given sections.
(i) We have to find the value of $\sqrt{x+1}\cot \phi $ . Let us express $\cot \phi $ in terms of the sides of a right-angled triangle for a specific angle $\phi $ .
We know that $\cot \phi =\dfrac{\text{Adjacent side}}{\text{Opposite side}}$ . Here, we can see that the adjacent side specific to $\phi $ is QR and the opposite side is PQ. Therefore, we can write x$\sqrt{x+1}\cot \phi $ as
$\Rightarrow \sqrt{x+1}\cot \phi =\sqrt{x+1}\times \dfrac{QR}{PQ}$
Let us substitute (i) and the value of QR in the above expression.
$\Rightarrow \sqrt{x+1}\cot \phi =\sqrt{x+1}\times \dfrac{x}{2\sqrt{x+1}}$
Let us cancel the common terms.
$\begin{align}
  & \Rightarrow \sqrt{x+1}\cot \phi =\require{cancel}\cancel{\sqrt{x+1}}\times \dfrac{x}{2\require{cancel}\cancel{\sqrt{x+1}}} \\
 & \Rightarrow \sqrt{x+1}\cot \phi =\dfrac{x}{2} \\
\end{align}$
Hence, the value of $\sqrt{x+1}\cot \phi $ is $\dfrac{x}{2}$ .
(ii) We have to find the value of $\sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta $ . We know that $\tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}$ .
From the figure, we can see that the opposite side specific to $\theta $ is QR and the adjacent side is PQ. Therefore, we can write $\sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta $ as
$\Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =\sqrt{{{x}^{3}}+{{x}^{2}}}\times \dfrac{QR}{PQ}$
Let us substitute (i) and the value of QR in the above expression.
$\Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =\sqrt{{{x}^{3}}+{{x}^{2}}}\times \dfrac{x}{2\sqrt{x+1}}$
Let us take ${{x}^{2}}$ outside from $\sqrt{{{x}^{3}}+{{x}^{2}}}$ .
$\begin{align}
  & \Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =\sqrt{{{x}^{2}}\left( x+1 \right)}\times \dfrac{x}{2\sqrt{x+1}} \\
 & \Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =x\sqrt{x+1}\times \dfrac{x}{2\sqrt{x+1}} \\
\end{align}$
Let us cancel the common terms.
$\begin{align}
  & \Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =x\require{cancel}\cancel{\sqrt{x+1}}\times \dfrac{x}{2\require{cancel}\cancel{\sqrt{x+1}}} \\
 & \Rightarrow \sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta =\dfrac{{{x}^{2}}}{2} \\
\end{align}$
Hence, the value of $\sqrt{{{x}^{3}}+{{x}^{2}}}\tan \theta $ is $\dfrac{{{x}^{2}}}{2}$ .
(iii) We have to find the value of $\cos \theta $ . We know that $\cos \theta =\dfrac{\text{Adjacent side}}{\text{Hypotenuse}}$ . From the given figure, we can see that the adjacent side specific to $\theta $ is PQ and the hypotenuse is PR.
$\begin{align}
  & \Rightarrow \cos \theta =\dfrac{PQ}{PR} \\
 & \Rightarrow \cos \theta =\dfrac{2\sqrt{x+1}}{x+2} \\
\end{align}$
Hence, the value of $\cos \theta $ is $\dfrac{2\sqrt{x+1}}{x+2}$ .

Note: Students must be thorough with the formulas of trigonometric ratios in terms of the sides of a right-angled triangle. They must note that we can apply these formulas and Pythagoras theorem only if the triangle is a right-angled one. We can find the adjacent and opposite sides as follows: