Ex.8.1 Q9 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

In the $$\Delta {ABC}$$ right-angled at $${B}$$, if \begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align} find the value of:

(i) $$\quad \text{sin }A \text{ cos }{C }+\text{ }\text{cos }A \;\text{sin }{C }$$

(ii) $$\quad \text{cos }{A}\;\text{cos }{C }-\text{ }\text{sin }{A }\text{ sin }{C }$$

Text Solution

Reasoning:

Using $$\tan {{A}} = \frac{1}{{\sqrt 3 }}$$, we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

(i) Let \begin{align}\triangle {ABC}\end{align} be a right-angled triangle \begin{align}\text{tan}\,A=\frac{\text{1}}{\sqrt{\text{3}}}\end{align}

\begin{align}\text{tan}\,A\,& =\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle A}{\text{side}\ \text{adjacent}\ \text{to}\ \angle A} \\ & =\frac{BC}{AB}=\frac{1}{\sqrt{3}}\end{align}

Let  $${BC} = {k}$$ and $${AB} = \sqrt{3\,}{k}$$  where $$k$$ is a positive real number.

By applying Pythagoras theorem for $$\text{ }\!\!\Delta\!\!\text{ }\,ABC$$

\begin{align} {AC}^{2} &={AB}^{2}+{BC}^{2} \\ &=(\sqrt{3} {k})+({k})^{2} \\ &=3 {k}^{2}+{k}^{2} \\ &=4 {k}^{2} \\ {AC} &=\sqrt{4 {k}^{2}} \\ &=2 {k} \end{align}

Therefore,

\begin{align}\sin A&=\frac{\text { side opposite to } \angle A}{\text { hypotenuse }} \\ &=\frac{B C}{A C}=\frac{1}{2} \\ \cos A&=\frac{\text { side adjacent to } \angle A}{\text { hypotenuse }} \\&=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \sin C&=\frac{\text { side opposite to } \angle C}{\text { hypotenuse }} \\ &=\frac{A B}{A C}=\frac{\sqrt{3}}{2} \\ \cos C&=\frac{\text { side adjacent to } \angle C}{\text { hypotenuse }} \\ & =\frac{B C}{A C}=\frac{1}{2}\end{align}

$$\text{(i) sin }A\text{ cos }C+\text{cos }A\text{ sin }C$$

(By substituting the values of the trigonometric functions in the above equation.)

\begin{align} & \text{ sin }A\text{ cos }C+\text{cos }A\text{ sin }C \\ &= \left( \frac{1}{2} \right)\! \left( \frac{1}{2} \right)\!+\!\left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right) \\ & =\frac{1}{4}+\frac{3}{4} \\ & =\frac{1+3}{4} \\ & =\frac{4}{4} \\ & =1 \end{align}

$$\text{(ii)}\cos \,A\,\cos \,C-\sin \,A\,\sin \,C$$

By substituting the values of the trigonometric functions in the above equation.

(ii)

\begin{align} & \cos \,A\,\cos \,C-\sin \,A\,\sin \,C \\ &= \left( \frac{\sqrt{3}}{2} \right)\! \left( \frac{1}{2} \right) \!-\!\left( \frac{1}{2} \right)\! \left( \frac{\sqrt{3}}{2} \right) \\ & =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4} \\ & =0 \end{align}

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