Question

note: triangle may not be drawn to scale. suppose a = 8 and b = 5. find an exact value or give least two decimal places: sin(a) = cos(a) = tan(a) = sec(a) = csc(a) = cot(a) = question help: message instructor

Explanation:

Step1: Find the hypotenuse c

By the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$. Given $a = 8$ and $b = 5$, then $c=\sqrt{8^{2}+5^{2}}=\sqrt{64 + 25}=\sqrt{89}$.

Step2: Calculate sin(A)

$\sin(A)=\frac{a}{c}=\frac{8}{\sqrt{89}}=\frac{8\sqrt{89}}{89}\approx0.85$.

Step3: Calculate cos(A)

$\cos(A)=\frac{b}{c}=\frac{5}{\sqrt{89}}=\frac{5\sqrt{89}}{89}\approx0.53$.

Step4: Calculate tan(A)

$\tan(A)=\frac{a}{b}=\frac{8}{5} = 1.6$.

Step5: Calculate sec(A)

$\sec(A)=\frac{c}{b}=\frac{\sqrt{89}}{5}\approx1.88$.

Step6: Calculate csc(A)

$\csc(A)=\frac{c}{a}=\frac{\sqrt{89}}{8}\approx1.18$.

Step7: Calculate cot(A)

$\cot(A)=\frac{b}{a}=\frac{5}{8}=0.625$.

Answer:

$\sin(A)=\frac{8\sqrt{89}}{89}\approx0.85$
$\cos(A)=\frac{5\sqrt{89}}{89}\approx0.53$
$\tan(A)=1.6$
$\sec(A)=\frac{\sqrt{89}}{5}\approx1.88$
$\csc(A)=\frac{\sqrt{89}}{8}\approx1.18$
$\cot(A)=0.625$