Ex.8.1 Q1 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

In  \(\,\Delta \rm ABC,\) right-angled at \(\rm{B}\), \(\rm AB = 24 \,\rm cm,\) \(\rm BC = 7\rm\, cm\), determine:

\(\begin{align} & \left( \text{i} \right)\quad\text{sin}\,\rm{A},\text{cos}\,\rm{A} \\ & \left( \text{ii} \right)\quad \text{sin }\rm{C},\text{cos}\,\rm{C}  \end{align}\)

Text Solution

What is the known?

Two sides of a right-angled triangle \(\Delta\!\!\text{ ABC}\)

What is the unknown?

Sine and cosine of angle \(\rm A\) and \(\rm C\).

Reasoning:

Applying Pythagoras theorem for  \(\Delta \text{ABC,}\) we can find hypotenuse (side \(\rm AC\)). Once hypotenuse is known, we can find sine and cosine angle using trigonometric ratios.

Steps:

\(\Delta \text{ABC,}\) we obtain.

\[\begin{align}\text{A}{{\text{C}}^{\text{2}}}\, &=\,\text{A}{{\text{B}}^{\text{2}}}\text{+B}{{\text{C}}^{\text{2}}} \\ &=\,{{\text{(24}\ \text{cm)}}^{\text{2}}}\,\text{+}\,{{\text{(7}\ \text{cm)}}^{\text{2}}} \\ &=\,\text{(576+49)}\ \text{c}{{\text{m}}^{\text{2}}} \\ &=\,\text{625}\ \text{c}{{\text{m}}^{\text{2}}} \end{align}\]

\(\therefore\) Hypotenuse \(\text{AC}\,\text{=}\,\sqrt{\text{625}}\ \text{cm}\,\text{=}\,\text{25}\ \text{cm}\)

(i)

\[\begin{align} & \sin \text{A}=\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}} \\ & \sin \text{A}=\frac{\text{7}\,\text{cm}}{\text{25}\,\text{cm}}=\frac{7}{25} \\ & \sin \text{A}=\frac{7}{25} \\ & \cos \text{A}=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}}=\frac{\text{AB}}{\text{AC}} \\ & \ \ \ \ \ \ \ \ =\frac{24\,\text{cm}}{25\,\text{cm}}=\frac{24}{25} \\ & \cos \text{A}=\frac{24}{25} \end{align}\]

(ii)

\[\begin{align} & \text{sin}\,\text{C=}\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}}\text{=}\frac{\text{AB}}{\text{AC}} \\ & \text{sin}\,\text{C=}\frac{24\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{24}{\text{25}} \\ & \text{sin}\,\text{C=}\frac{\text{24}}{\text{25}} \\ & \text{cos}\,\text{C=}\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{C}}{\text{hypotenuse}}\text{=}\frac{\text{BC}}{\text{AC}} \\ & \ \ \ \ \ \ \ \,\,\text{=}\frac{7\,\text{cm}}{\text{25}\,\text{cm}}\text{=}\frac{7}{\text{25}} \\ & \text{cos}\,\text{C=}\frac{7}{\text{25}} \end{align}\]