QuestionMay 26, 2026

Find the angle beta between 0^circ and 180^circ that satisfies the following equation. 21^2=20^2+29^2-(2)(20)(29)cosbeta beta approx square ^circ (Round to one decimal place as needed.)

Find the angle beta between 0^circ and 180^circ that satisfies the following equation. 21^2=20^2+29^2-(2)(20)(29)cosbeta beta approx square ^circ (Round to one decimal place as needed.)
Find the angle beta  between 0^circ  and 180^circ  that satisfies the following equation.
21^2=20^2+29^2-(2)(20)(29)cosbeta 
beta approx square ^circ 
(Round to one decimal place as needed.)

Solution
4.4(186 votes)

Answer

46.4^{\circ} Explanation 1. Isolate the cosine term Rearrange the equation 21^{2}=20^{2}+29^{2}-(2)(20)(29)\cos\beta to solve for \cos\beta using the **Law of Cosines**: c^2 = a^2 + b^2 - 2ab\cos C. 441 = 400 + 841 - 1160\cos\beta 441 = 1241 - 1160\cos\beta -800 = -1160\cos\beta 2. Solve for \cos\beta Divide both sides by -1160. \cos\beta = \frac{800}{1160} = \frac{80}{116} = \frac{20}{29} 3. Calculate the angle \beta Apply the inverse cosine function: \beta = \arccos(\frac{20}{29}). \beta \approx 46.397^{\circ}

Explanation

1. Isolate the cosine term<br />Rearrange the equation $21^{2}=20^{2}+29^{2}-(2)(20)(29)\cos\beta$ to solve for $\cos\beta$ using the **Law of Cosines**: $c^2 = a^2 + b^2 - 2ab\cos C$.<br />$441 = 400 + 841 - 1160\cos\beta$<br />$441 = 1241 - 1160\cos\beta$<br />$-800 = -1160\cos\beta$<br /><br />2. Solve for $\cos\beta$<br />Divide both sides by $-1160$.<br />$\cos\beta = \frac{800}{1160} = \frac{80}{116} = \frac{20}{29}$<br /><br />3. Calculate the angle $\beta$<br />Apply the inverse cosine function: $\beta = \arccos(\frac{20}{29})$.<br />$\beta \approx 46.397^{\circ}$
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