Question

In a right ∆ ABC , right angled at B , if sin A=3/5, find all the six trigonometric ratios of angle C

Answer 1

$★ {\underline{\sf\large{ Concept }}}$

The Trigonometric ratios depend only on the value of angle ∅ and are independent of the position of the point P on the terminal side XY of the acute angle $\angle$ZXY.

The Trigonometric ratio are same for the same angle. We have to prove that the Trigonometric ratios of angle ∅ are same in both the Triangles.

$★ {\pmb{\underline{\sf{ Explained... }}}}$

As We know that In right angled ∆ABC right angled at B we have,

$\circ \ {\underline{\boxed{\sf\gray{ Sin \ \theta = \dfrac{Perpendicular}{Hypotenuse} }}}} \\ \\ \colon\implies {\sf{ Sin \ A = \dfrac{3}{5} }}$

Now, We draw a right angled ∆ABC right angled at B such that,

• Perpendicular {BC} = 3
• Hypotenuse {AC} = 5

So, By applying Pythagoras theorem, we can find the value of third side {AB} as:

$\colon\implies{\sf{ AC^2 = AB^2 + BC^2 }} \\ \\ \colon\implies{\sf{ 5^2 = AB^2 + 3^2 }} \\ \\ \colon\implies{\sf{ 25 = AB^2 + 9 }} \\ \\ \colon\implies{\sf{ AB^2 = 25-9 }} \\ \\ \colon\implies{\sf{ AB = \pm \sqrt{16} }} \\ \\ \colon\implies{\sf{ AB = \pm 4_{(Base)} }}$

Now Finally, We've:

• Base [AB] = 4
• Perpendicular [BC] = 3
• Hypotenuse [AC] = 5

We have to consider the T-Ratios of $\angle$C.

• Base [BC] = 3
• Perpendicular [AB] = 4
• Hypotenuse [AC] = 5

$\circ \ {\underline{\boxed{\sf{ Sin \ C = \dfrac{Perpendicular}{Hypotenuse} = \dfrac{4}{5} }}}} \\ \\ \circ \ {\underline{\boxed{\sf{ Cos \ C = \dfrac{Base}{Hypotenuse} = \dfrac{3}{5} }}}} \\ \\ \circ \ {\underline{\boxed{\sf{ tan \ C = \dfrac{Perpendicular}{Base} = \dfrac{4}{3} }}}} \\ \\ \circ \ {\underline{\boxed{\sf{ Cot \ C = \dfrac{Base}{Perpendicular} = \dfrac{3}{4} }}}} \\ \\ \circ \ {\underline{\boxed{\sf{ Sec \ C = \dfrac{Hypotenuse}{Base} = \dfrac{5}{3} }}}} \\ \\ \circ \ {\underline{\boxed{\sf{ Cosec \ C = \dfrac{Hypotenuse}{Perpendicular} = \dfrac{5}{4} }}}}$

[PROVED]