Answer:
See explanation
Step-by-step explanation:
Use trigonometric functions to solve your problems:
[tex]\cos \theta=\dfrac{\text{Adjacent leg}}{\text{Hypotenuse}}\\ \\\sin \theta=\dfrac{\text{Opposite leg}}{\text{Hypotenuse}}\\ \\\tan \theta=\dfrac{\text{Opposite leg}}{\text{Adjacent leg}}\\ \\\cot \theta=\dfrac{\text{Adjacent leg}}{\text{Opposite leg}}[/tex]
Q1. From the diagram,
Adjacent leg AC = 12
Hypotenuse AB = 13
Then [tex]\cos \theta =\dfrac{12}{13}\Rightarrow \theta =\arccos \dfrac{12}{13}\approx 22.6^{\circ}[/tex]
Q2. From the diagram,
Adjacent leg AC = 13
Opposite leg BC = 4
Then [tex]\tan \theta =\dfrac{4}{13}\Rightarrow \theta =\arctan \dfrac{4}{13}\approx 17.1^{\circ}[/tex]
Q3. From the diagram,
Adjacent leg AC = 6
Hypotenuse AB = 9
Then [tex]\cos \theta =\dfrac{6}{9}\Rightarrow \theta =\arccosn \dfrac{6}{9}\approx 48.2^{\circ}[/tex]
Q4. From the diagram,
Adjacent leg AC = 10
Opposite leg BC = 11.9
Then [tex]\tan \theta =\dfrac{11.9}{10}\Rightarrow \theta =\arctan \dfrac{11.9}{10}\approx 50^{\circ}[/tex]
Q5. From the diagram,
Adjacent leg BC = 14
Opposite leg AC = 7.7
Then [tex]\tan \theta =\dfrac{7.7}{14}\Rightarrow \theta =\arctan \dfrac{7.7}{14}\approx 28.8^{\circ}[/tex]
Q6. From the diagram,
Adjacent leg BC = 4
Hypotenuse AB = 5
Then [tex]\cos \theta =\dfrac{4}{5}\Rightarrow \theta =\arccosn \dfrac{4}{5}\approx 36.9^{\circ}[/tex]
Q7. From the diagram,
Adjacent leg BC = 4.4
Hypotenuse AB = 11
Then [tex]\cos \theta =\dfrac{4.4}{11}\Rightarrow \theta =\arccosn \dfrac{4.4}{11}\approx 66.4^{\circ}[/tex]
