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Q1(ii):
In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (ii) $\sin C, \cos C$

Solution :

Given: A right-angled triangle $ABC$ where $\angle B = 90^\circ$, side $AB = 24$ cm, and side $BC = 7$ cm.

To find: The values of $\sin C$ and $\cos C$.

B C A 7 cm 24 cm

Step 1: Calculating the Hypotenuse ($AC$)

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides [Pythagoras Theorem].

$AC^2 = AB^2 + BC^2$

Substitute the given values:

$AC^2 = (24)^2 + (7)^2$

$AC^2 = 576 + 49$

$AC^2 = 625$

$AC = \sqrt{625} = 25$ cm

Step 2: Defining Trigonometric Ratios for $\angle C$

For $\angle C$, the side opposite is $AB$ and the side adjacent is $BC$. The hypotenuse is $AC$.

$\sin C = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC}$

$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC}$

Step 3: Substituting the values

Using $AB = 24$ cm, $BC = 7$ cm, and $AC = 25$ cm:

$\sin C = \frac{24}{25}$

$\cos C = \frac{7}{25}$

Final Answer: $\sin C = \frac{24}{25}$ and $\cos C = \frac{7}{25}$


More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1


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