Question

trigonometric ratios maze! directions: beginning at the start box, find the indicated trigonometric ratio. use your simplified solutions to navigate through the maze. version 1: pythagorean triples only @ gina wilson (all things algebra®. llc), 2023

Explanation:

Step1: Calculate the hypotenuse of the starting triangle

In the starting triangle (with sides \(x = 16\), \(y = 30\)), using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 16\), \(b = 30\). Then \(c=\sqrt{16^{2}+30^{2}}=\sqrt{256 + 900}=\sqrt{1156}=34\).

Step2: Calculate \(\sin Z\)

The formula for \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle Z\), the opposite side to \(\angle Z\) is \(x = 16\), and the hypotenuse \(c = 34\). So \(\sin Z=\frac{16}{34}=\frac{8}{17}\).

Step3: Navigate to the next box

Since \(\sin Z=\frac{8}{17}\), we move to the box labeled “Find: \(\cos A\)” (because the value \(\frac{8}{17}\) is connected to it in the maze - like structure).

Step4: Calculate the third side of the triangle for \(\cos A\)

In the triangle for \(\cos A\) (with hypotenuse \(25\) and one side \(7\)), using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), let \(b = 7\), \(c = 25\), then \(a=\sqrt{25^{2}-7^{2}}=\sqrt{625 - 49}=\sqrt{576}=24\).

Step5: Calculate \(\cos A\)

The formula for \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle A\), the adjacent side is \(24\), and the hypotenuse is \(25\). So \(\cos A=\frac{24}{25}\).

Step6: Navigate to the next box

Since \(\cos A=\frac{24}{25}\), we move to the box labeled “Find: \(\tan T\)” (as \(\frac{24}{25}\) is connected to it).

Step7: Calculate \(\tan T\)

In the triangle for \(\tan T\) (with sides \(9\) and \(15\)), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle T\), \(\tan T=\frac{15}{9}=\frac{5}{3}\).

Step8: Navigate to the next box

Since \(\tan T=\frac{5}{3}\), we move to the box labeled “Find: \(\tan A\)” (as \(\frac{5}{3}\) is connected to it).

Step9: Calculate \(\tan A\)

In the triangle for \(\tan A\) (with sides \(4\) and \(3\) - since \(BC=\sqrt{5^{2}-4^{2}} = 3\)), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle A\), \(\tan A=\frac{3}{4}\).

Step10: Navigate to the next box

Since \(\tan A=\frac{3}{4}\), we move to the box labeled “Find: \(\cos L\)” (as \(\frac{3}{4}\) is connected to it).

Step11: Calculate the hypotenuse for \(\cos L\)

In the triangle for \(\cos L\) (with sides \(32\) and \(68\)), using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), let \(a = 32\), \(b=60\) (since \(68^{2}-32^{2}=(68 + 32)(68 - 32)=100\times36\), so the other side is \(60\)), \(c = 68\). \(\cos L=\frac{60}{68}=\frac{15}{17}\).

Step12: Navigate to the next box

Since \(\cos L=\frac{15}{17}\), we move to the box labeled “Find: \(\tan Z\)” (as \(\frac{15}{17}\) is connected to it).

Step13: Calculate \(\tan Z\)

In the triangle for \(\tan Z\) (with sides \(60\) and \(87\) - \(YZ^{2}-XY^{2}=87^{2}-60^{2}=(87 + 60)(87 - 60)=147\times27\), the other side \(XZ = 63\)), \(\tan Z=\frac{60}{63}=\frac{20}{21}\).

Step14: Navigate to the next box

Since \(\tan Z=\frac{20}{21}\), we move to the box labeled “Find: \(\cos S\)” (as \(\frac{20}{21}\) is connected to it).

Step15: Calculate \(\cos S\)

In the triangle for \(\cos S\) (with hypotenuse \(50\) and one side \(48\)), using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), let \(b = 48\), \(c = 50\), then \(a=\sqrt{50^{2}-48^{2}}=\sqrt{(50 + 48)(50 - 48)}=\sqrt{98\times2}=\sqrt{196}=14\). \(\cos S=\frac{14}{50}=\frac{7}{25}\).

Step16: Navigate to the next box

Since \(\cos S=\frac{7}{25}\), we move to the box labeled “Find: \(\cos R\)” (as \(\frac{7}{25}\) is connected to it).

Step17: Calculate \(\cos R\)

In the triangle for \(\cos R\) (with hypotenuse \(52\) - \(48^{2}+20^{2}=2304+400 = 2704=52^{2}\)), \(\cos R=\frac{48}{52}=\frac{12}{13}\).

Step18: Navigate to the “End” box

Since \(\frac{12}{13}\) is connected to the “End” box.

Answer:

The path through the maze is: Start (calculate \(\sin Z=\frac{8}{17}\)) → Find: \(\cos A\) (\(\frac{24}{25}\)) → Find: \(\tan T\) (\(\frac{5}{3}\)) → Find: \(\tan A\) (\(\frac{3}{4}\)) → Find: \(\cos L\) (\(\frac{15}{17}\)) → Find: \(\tan Z\) (\(\frac{20}{21}\)) → Find: \(\cos S\) (\(\frac{7}{25}\)) → Find: \(\cos R\) (\(\frac{12}{13}\)) → End.