If sin θ = (9/41), 0° < θ < 90° then what is the value of cot θ ?
40/9
The question asks us to find the value of cot θ given that sin θ = \(\frac{9}{41}\) and the angle θ is in the first quadrant (\(0° < \theta < 90°\)).
In trigonometry, the basic ratios relate the angles of a right-angled triangle to the lengths of its sides.
Given sin θ = \(\frac{9}{41}\), we can represent this using a right-angled triangle where:
To find cot θ, we need the length of the adjacent side. We can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (opposite and adjacent).
Let the adjacent side be represented by \(x\). According to the Pythagorean theorem:
\((\text{Opposite})^2 + (\text{Adjacent})^2 = (\text{Hypotenuse})^2\)
\((9)^2 + (x)^2 = (41)^2\)
\(81 + x^2 = 1681\)
Now, we solve for \(x\):
\(x^2 = 1681 - 81\)
\(x^2 = 1600\)
\(x = \sqrt{1600}\)
\(x = 40\)
So, the length of the adjacent side is 40 units.
Now we have all the side lengths needed to find cot θ:
The formula for cot θ is:
\(\cot \theta = \frac{\text{Adjacent}}{\text{Opposite}}\)
Substituting the values:
\(\cot \theta = \frac{40}{9}\)
The problem specifies that \(0° < \theta < 90°\), which means θ is in the first quadrant. In the first quadrant, all trigonometric ratios (sin, cos, tan, cot, sec, cosec) are positive. Our calculated value \(\frac{40}{9}\) is positive, which is consistent with the given range of θ.
Thus, the value of cot θ is \(\frac{40}{9}\).
| Side | Length |
|---|---|
| Opposite | 9 |
| Hypotenuse | 41 |
| Adjacent | 40 |
| Ratio | Definition (Sides) | Relationship to other ratios |
|---|---|---|
| sin θ | \(\frac{\text{Opposite}}{\text{Hypotenuse}}\) | \(\frac{1}{\csc \theta}\) |
| cos θ | \(\frac{\text{Adjacent}}{\text{Hypotenuse}}\) | \(\frac{1}{\sec \theta}\) |
| tan θ | \(\frac{\text{Opposite}}{\text{Adjacent}}\) | \(\frac{1}{\cot \theta}, \frac{\sin \theta}{\cos \theta}\) |
| cot θ | \(\frac{\text{Adjacent}}{\text{Opposite}}\) | \(\frac{1}{\tan \theta}, \frac{\cos \theta}{\sin \theta}\) |
| sec θ | \(\frac{\text{Hypotenuse}}{\text{Adjacent}}\) | \(\frac{1}{\cos \theta}\) |
| csc θ | \(\frac{\text{Hypotenuse}}{\text{Opposite}}\) | \(\frac{1}{\sin \theta}\) |
Besides using the sides of a right triangle, we can also use trigonometric identities to solve such problems. For example, we know that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).
If we have sin θ, we can find cos θ using the identity \(\sin^2 \theta + \cos^2 \theta = 1\).
\(\left(\frac{9}{41}\right)^2 + \cos^2 \theta = 1\)
\(\frac{81}{1681} + \cos^2 \theta = 1\)
\(\cos^2 \theta = 1 - \frac{81}{1681} = \frac{1681 - 81}{1681} = \frac{1600}{1681}\)
\(\cos \theta = \pm \sqrt{\frac{1600}{1681}} = \pm \frac{40}{41}\)
Since \(0° < \theta < 90°\), θ is in the first quadrant, where cos θ is positive. So, cos θ = \(\frac{40}{41}\).
Now we can find cot θ:
\(\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{41} \times \frac{41}{9} = \frac{40}{9}\)
This confirms the result obtained using the right triangle method. Understanding the quadrant of the angle is crucial for determining the sign of trigonometric ratios. In the first quadrant (0° to 90°), all ratios are positive.
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