The measurement of one side of a right triangle is found to be 9.5 inches, and the angle opposite that slde is with a possible error of . (Round your answers
to two decimal places.)
(a) Approximate the percent error (in radians) in computing the length of the hypotenuso.
(b) Estimate the maximum allowable percent error (in radlans) in measuring the angle if the error in computing the length of the hypctenuse cannot exceed .

Ask by Bernard King. in the United States

Mar 31,2025

Question

The measurement of one side of a right triangle is found to be 9.5 inches, and the angle opposite that slde is with a possible error of . (Round your answers
to two decimal places.)
(a) Approximate the percent error (in radians) in computing the length of the hypotenuso.
(b) Estimate the maximum allowable percent error (in radlans) in measuring the angle if the error in computing the length of the hypctenuse cannot exceed .

Ask by Bernard King. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We begin by noting that the side of length 9.5 inches is opposite an angle $$\theta$$ whose measured value is
$$
\theta = 26^\circ 45' = 26 + \frac{45}{60} = 26.75^\circ.
$$
In radians this is
$$
\theta \approx 26.75 \times \frac{\pi}{180} \approx 0.466\, \text{radians}.
$$

Since 9.5 inches is the side opposite $$\theta$$, the hypotenuse $$h$$ is given by
$$
h = \frac{9.5}{\sin \theta}.
$$

Any error in $$\theta$$ will affect the computed $$h$$. We use differential (or linearized) error analysis.

The derivative of $$h$$ with respect to $$\theta$$ is
$$
\frac{dh}{d\theta} = -\frac{9.5\,\cos \theta}{\sin^2 \theta}.
$$
Dividing by $$h$$ we obtain the relative error in $$h$$:
$$
\frac{dh}{h} = -\frac{9.5\,\cos \theta}{\sin^2 \theta}\bigg/\frac{9.5}{\sin \theta}\, d\theta = -\cot\theta\, d\theta.
$$
Thus, the magnitude of the relative error is
$$
\left|\frac{dh}{h}\right| = |\cot\theta|\, |d\theta|.
$$

We now address each part.

---

**(a) Percent error in computing the hypotenuse**

The given possible error in measuring the angle is $$15'$$. In degrees,
$$
15' = \frac{15}{60} = 0.25^\circ.
$$
In radians the error is
$$
d\theta \approx 0.25 \times \frac{\pi}{180} \approx 0.00436\,\text{radians}.
$$

Next, we compute $$\cot\theta$$. With $$\theta \approx 0.466\,\text{radians}$$,
$$
\sin(0.466) \approx 0.451,\quad \cos(0.466) \approx 0.893,
$$
so that
$$
\cot(0.466) = \frac{\cos(0.466)}{\sin(0.466)} \approx \frac{0.893}{0.451} \approx 1.98.
$$

The relative error in $$h$$ is then
$$
\left|\frac{dh}{h}\right| \approx 1.98 \times 0.00436 \approx 0.00863.
$$
Expressed as a percent, the error in $$h$$ is
$$
0.00863 \times 100\% \approx 0.86\%.
$$

---

**(b) Maximum allowable percent error in measuring the angle**

Now suppose that we require the error in the computed hypotenuse not to exceed $$5\%$$. This means that
$$
\left|\frac{dh}{h}\right| \le 0.05.
$$
Since
$$
\left|\frac{dh}{h}\right| \approx |\cot\theta|\,|d\theta|,
$$
we have
$$
|\cot\theta|\,|d\theta| \le 0.05.
$$
Solving for $$d\theta$$,
$$
|d\theta| \le \frac{0.05}{|\cot\theta|} \approx \frac{0.05}{1.98} \approx 0.0253\,\text{radians}.
$$

Thus, the maximum allowable error in $$\theta$$ is about $$0.0253$$ radians. To express this as a percent error relative to the measured angle $$\theta \approx 0.466$$ radians, we have
$$
\text{Percent error in } \theta \approx \frac{0.0253}{0.466} \times 100\% \approx 5.43\%.
$$

---

**Answers**

(a) The percent error in computing the hypotenuse is approximately $$0.86\%$$.

(b) To keep the error in the hypotenuse below $$5\%$$, the maximum allowable error in measuring $$\theta$$ is about $$0.0253$$ radians (which is approximately a $$5.43\%$$ error in $$\theta$$).
(a) The percent error in computing the hypotenuse is approximately 0.86%. (b) To keep the error in the hypotenuse below 5%, the maximum allowable error in measuring the angle is about 0.0253 radians, which is approximately a 5.43% error in the angle.

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