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

Go back to  'Ex.8.1'

## 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) $$\quad \text{sin }C,\text{cos}\,C$$

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 1

## 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}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school