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

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Question

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

(i) \(  \quad\text{sin}\,A,\text{cos}A \)

(ii) \(\quad \text{sin }C,\text{cos}\,C  \)

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 \( A\) and \( C\).

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)

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