Question

note: triangle may not be drawn to scale. suppose a = 9 and b = 4. round answers to one decimal place: sin(a) = cos(a) = tan(a) = sec(a) = csc(a) = cot(a) = question help: video

Explanation:

Step1: Calculate hypotenuse $c$

Use Pythagorean theorem:
$$c = \sqrt{a^2 + b^2} = \sqrt{9^2 + 4^2} = \sqrt{81 + 16} = \sqrt{97} \approx 9.8$$

Step2: Compute $\sin(A)$

$\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c}$
$\sin(A) = \frac{9}{\sqrt{97}} \approx 0.9$

Step3: Compute $\cos(A)$

$\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c}$
$\cos(A) = \frac{4}{\sqrt{97}} \approx 0.4$

Step4: Compute $\tan(A)$

$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}$
$\tan(A) = \frac{9}{4} = 2.3$

Step5: Compute $\sec(A)$

$\sec(A) = \frac{1}{\cos(A)} = \frac{c}{b}$
$\sec(A) = \frac{\sqrt{97}}{4} \approx 2.5$

Step6: Compute $\csc(A)$

$\csc(A) = \frac{1}{\sin(A)} = \frac{c}{a}$
$\csc(A) = \frac{\sqrt{97}}{9} \approx 1.1$

Step7: Compute $\cot(A)$

$\cot(A) = \frac{1}{\tan(A)} = \frac{b}{a}$
$\cot(A) = \frac{4}{9} \approx 0.4$

Answer:

$\sin(A) = 0.9$
$\cos(A) = 0.4$
$\tan(A) = 2.3$
$\sec(A) = 2.5$
$\csc(A) = 1.1$
$\cot(A) = 0.4$