Question

If 3 tanθ =4. Find all the trygnometric ratio​

Answer 1

Question:

If 3 tanθ =4. Find all the trignometric ratios.

Solution:

$3tanθ = 4 \\ tanθ = \frac{4}{3} = \frac{opposite \: side}{adjacent \: side} \\ In \: \triangle ABC \: by , \: Pythagoras \: theorem \\ AB^2+BC^2=AC^2 \: \\ {(4)}^{2} + {(3)}^{2} = AC^2 \\ AC^2 = 16 + 9 \\ AC^2 = 25 \\ AC= 5$

From Figure,

$sinθ = \frac{opposite \: side}{hypotenuse} = \frac{4}{5} \\ cosθ = \frac{adjacent \: side}{hypotenuse} = \frac{3}{5} \\ cosecθ = \frac{hypotenuse}{opposite \: side} = \frac{5}{4} \\ secθ = \frac{hypotenuse}{adjacent \: side} = \frac{5}{3} \\ cotθ = \frac{adjacent \: side}{opposite \: side} = \frac{3}{4}$

_______________________

Answer 2

ANSWER:-

Given:

If 3 tan Ф= 4.

To find:

All the trignometric  ratio.

Explanation:

tanФ= $\frac{4}{3}$

tanФ= $\frac{Perpendicular}{Base}$

• Perpendicular= 4
• Base= 3
• Hypotenuse= ?

Using Pythagoras Theorem:

[Hypotenuse]² = [Base]² + [Perpendicular]²

[Hypotenuse]²= (3)² + (4)²

[Hypotenuse]²= 9 + 16

[Hypotenuse]²= 25

Hypotenuse= √25

Hypotenuse= 5

Now,

All the trignometric ratio:

• sinФ= $\frac{P}{H} = \frac{4}{5}$
• cosФ= $\frac{B}{H}= \frac{3}{5}$
• tanФ= $\frac{P}{B} = \frac{4}{3}$
• cotФ= $\frac{B}{P} =\frac{3}{4}$
• secФ= $\frac{H}{B} = \frac{5}{3}$
• cosecФ= $\frac{H}{P} = \frac{5}{4}$

Note:

The value of sinФ, cosФ, tanФ etc, depends on the angleФ, not on the size of the right angled Δ.