Question

evaluate to the nearest tenth: cot 112°5= -2.5 1.3 0.9 -0.4 question 18 1 pts find the value of x, to the nearest tenth, in the right triangle below. 41 13 x

Explanation:

Step1: Convert the angle from degrees and minutes to decimal degrees

Since \(1^{\circ}=60'\), then \(5'=\frac{5}{60}\approx0.083^{\circ}\). So \(112^{\circ}5'\approx112.083^{\circ}\).

Step2: Use the cotangent identity \(\cot\theta=\frac{\cos\theta}{\sin\theta}\)

\(\cot(112.083^{\circ})=\frac{\cos(112.083^{\circ})}{\sin(112.083^{\circ})}\)
We know that \(\cos(112.083^{\circ})\approx - 0.375\) and \(\sin(112.083^{\circ})\approx0.927\)
\(\cot(112.083^{\circ})=\frac{-0.375}{0.927}\approx - 0.4\)

Answer:

-0.4