Find a value of θ, in the interval [0°, 90°) that satisfies each statement. Give answers in decimal degrees to six decimal places. sec θ = 1.2637891
Ch. 2 - Acute Angles and Right Triangles
Chapter 3, Problem 2.R.32
Find a value of θ, in the interval [0°, 90°) that satisfies each statement. Give answers in decimal degrees to six decimal places. cot θ = 1.1249386
Verified step by step guidance1
Recall the definition of cotangent in terms of tangent: \(\cot \theta = \frac{1}{\tan \theta}\). This means \(\tan \theta = \frac{1}{\cot \theta}\).
Calculate \(\tan \theta\) by taking the reciprocal of the given cotangent value: \(\tan \theta = \frac{1}{1.1249386}\).
Use the inverse tangent function to find \(\theta\): \(\theta = \tan^{-1} \left( \frac{1}{1.1249386} \right)\).
Make sure your calculator is set to degree mode since the problem asks for the angle in degrees.
Evaluate the inverse tangent expression to find \(\theta\) in degrees, and round your answer to six decimal places.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Cotangent Function
The cotangent of an angle θ in a right triangle is the ratio of the adjacent side to the opposite side, or equivalently, cot θ = 1 / tan θ. It is the reciprocal of the tangent function and is used to relate angles to side lengths in trigonometry.
Recommended video:
Introduction to Cotangent Graph
Inverse Trigonometric Functions
Inverse trigonometric functions, such as arccotangent, allow us to find an angle when given a trigonometric ratio. Since cotangent is less common, the angle θ can be found by taking the arctangent of the reciprocal of the given cotangent value.
Recommended video:
Introduction to Inverse Trig Functions
Angle Measurement and Interval Restrictions
Angles are often measured in degrees within specified intervals. Here, θ must lie in [0°, 90°), meaning the solution is an acute angle. This restriction ensures the angle corresponds to the first quadrant where cotangent values are positive.
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Reference Angles on the Unit Circle
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