Question

In $△ABC,$ sin C = $\frac{3}{5}.$ Match the given trigonometric ratios with their values.

• A. $\frac{4}{5}$
• B. $\frac{3}{4}$
• C. $\frac{5}{3}$
• D. $\frac{5}{4}$

Answer 1

It is given that sin C = $\frac{3}{5}.$
But, sin C = $\frac{\text{Opposite side(AB)}}{\text{Hypotenuse (AC)}}.$
Thus, we have AB = 3k and AC = 5k, where k is a natural number.

From Pythagoras' theorem, we have
BC = $\sqrt{(5k{)}^2-(3k{)}^2}$
= $\sqrt{25k^2-9k^2}$
= $\sqrt{16k^2}$
= 4k cm

Now, the remaining ratios can be calculated as follows:

cos C = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{4}{5}$
tan C = $\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{3}{4}$
cosec C = $\frac{1}{\text{sin}C}=\frac{5}{3}$
sec C = $\frac{1}{\text{cos}C}=\frac{5}{4}$