Question

find the value of \\(\sin i\\) rounded to the nearest hundredth, if necessary.
answer
attempt 1 out of 2
\\(\sin i = \boxed{}
\\)
submit answer

Explanation:

Step1: Find side JK via Pythagoras

Let $JK = x$. For right $\triangle IJK$:
$$x^2 + (\sqrt{70})^2 = 9^2$$
$$x^2 + 70 = 81$$
$$x^2 = 11 \implies x = \sqrt{11}$$

Step2: Define sine of angle $I$

$\sin I = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{JK}{IK}$
$$\sin I = \frac{\sqrt{11}}{9}$$

Step3: Calculate decimal value

$\sqrt{11} \approx 3.3166$, so:
$$\sin I \approx \frac{3.3166}{9} \approx 0.3685$$

Step4: Round to nearest hundredth

Round 0.3685 to two decimal places.

Answer:

0.37