Given 15 cot A = 8, find sin A and sec A.
sin A = \(\dfrac{15}{17}\), sec A = \(\dfrac{17}{8}\)
Given: \(15\cot A = 8\).
To find: \(\sin A\) and \(\sec A\).
Step 1: First isolate \(\cot A\)
\(15\cot A = 8\Rightarrow \cot A = \dfrac{8}{15}\).
Step 2: Convert \(\cot A\) into a right-triangle ratio
We know:
\(\cot A = \dfrac{\text{adjacent}}{\text{opposite}}\).
So we can assume a right triangle for angle \(A\) such that:
Adjacent side \(= 8\) units and Opposite side \(= 15\) units.
Student Note: In trigonometry, whenever you get a ratio like \(\cot A = \dfrac{8}{15}\), you can safely take the sides in the same ratio (8 and 15). This makes finding other ratios easy.
Step 3: Find the hypotenuse using Pythagoras theorem
Hypotenuse \(= \sqrt{8^2 + 15^2}\)
\(= \sqrt{64 + 225} = \sqrt{289} = 17\).
So, Hypotenuse \(= 17\) units.
Step 4: Find \(\sin A\)
\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{15}{17}\).
Step 5: Find \(\sec A\)
First recall:
\(\sec A = \dfrac{1}{\cos A}\) and \(\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}\).
So, \(\cos A = \dfrac{8}{17}\).
Therefore,
\(\sec A = \dfrac{1}{\cos A} = \dfrac{1}{\frac{8}{17}} = \dfrac{17}{8}\).
Final Answer: \(\sin A = \dfrac{15}{17}\), \(\sec A = \dfrac{17}{8}\).
Quick Check (optional): Since \(\cot A = \dfrac{\cos A}{\sin A}\), we get \(\dfrac{\cos A}{\sin A} = \dfrac{\frac{8}{17}}{\frac{15}{17}} = \dfrac{8}{15}\), which matches the given value. So the result is correct.