Subjects trigonometry

Inverse Cosine Angles

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Inverse Cosine Angles


1. **Problem Statement:** Calculate the inverse cosine (arccos) values and angles for given cosine values and right triangles. 2. **Formula and Rules:** - The inverse cosine function, $\cos^{-1}(x)$, gives the angle whose cosine is $x$. - For a right triangle, $\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}$. - Angles are measured in degrees. --- **Question 12: Calculate $\cos^{-1}$ values** 12a. $\cos^{-1}(0.725)$ - Use a calculator: $\cos^{-1}(0.725) \approx 43.07^\circ$ 12b. $\cos^{-1}(0.325)$ - $\cos^{-1}(0.325) \approx 71.06^\circ$ 12c. $\cos^{-1}(\frac{3}{7})$ - Calculate $\frac{3}{7} \approx 0.4286$ - $\cos^{-1}(0.4286) \approx 64.62^\circ$ 12d. $\cos^{-1}(\frac{5}{12})$ - Calculate $\frac{5}{12} \approx 0.4167$ - $\cos^{-1}(0.4167) \approx 65.37^\circ$ --- **Question 13: Calculate angle $\angle E$ given $\cos E$** 13a. $\cos E = 0.625$ - $E = \cos^{-1}(0.625) \approx 51.32^\circ \to 51^\circ$ 13b. $\cos E = 0.812$ - $E = \cos^{-1}(0.812) \approx 35.71^\circ \to 36^\circ$ 13c. $\cos E = \frac{3}{5} = 0.6$ - $E = \cos^{-1}(0.6) \approx 53.13^\circ \to 53^\circ$ 13d. $\cos E = \frac{7}{11} \approx 0.6364$ - $E = \cos^{-1}(0.6364) \approx 50.46^\circ \to 50^\circ$ --- **Question 14: Calculate $\angle E$ in right triangles** Use $\cos E = \frac{\text{adjacent side}}{\text{hypotenuse}}$ where $E$ is the angle at $E$. 14a. Triangle EJI, right angle at J - $EJ = 2.1$ cm (adjacent), $EI = 3.7$ cm (hypotenuse) - $\cos E = \frac{2.1}{3.7} \approx 0.5676$ - $E = \cos^{-1}(0.5676) \approx 55.44^\circ$ 14b. Triangle EFG, right angle at F - $EF = 10$ cm (adjacent), $EG = \sqrt{EF^2 + FG^2} = \sqrt{10^2 + 18^2} = \sqrt{100 + 324} = \sqrt{424} \approx 20.59$ cm (hypotenuse) - $\cos E = \frac{10}{20.59} \approx 0.4854$ - $E = \cos^{-1}(0.4854) \approx 61.04^\circ$ 14c. Triangle ACE, right angle at C - $AC = 4$ m (adjacent), $AE = 9$ m (hypotenuse) - $\cos E = \frac{4}{9} \approx 0.4444$ - $E = \cos^{-1}(0.4444) \approx 63.26^\circ$ 14d. Triangle EML, right angle at M - $EM = 13.4$ cm (adjacent), $EL = 16.1$ cm (hypotenuse) - $\cos E = \frac{13.4}{16.1} \approx 0.8323$ - $E = \cos^{-1}(0.8323) \approx 33.56^\circ$