Question

find ik.

write your answer as an integer or as a decimal rounded to the nearest tenth.
ik =
submit

Explanation:

Step1: Identify the trigonometric ratio

In right triangle \( KJI \), we know angle \( K = 24^\circ \), adjacent side to angle \( K \) is \( KJ = 9 \), and we need to find the hypotenuse \( IK \). The cosine function relates the adjacent side and the hypotenuse: \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \). So, \( \cos(24^\circ)=\frac{KJ}{IK} \).

Step2: Solve for \( IK \)

We can rearrange the formula to solve for \( IK \): \( IK = \frac{KJ}{\cos(24^\circ)} \). Substituting \( KJ = 9 \) and \( \cos(24^\circ) \approx 0.9135 \), we get \( IK=\frac{9}{0.9135} \approx 9.85 \). Rounding to the nearest tenth, \( IK \approx 9.9 \). Wait, no, wait. Wait, actually, in the triangle, angle at \( J \) is right angle, so \( \angle J = 90^\circ \), \( KJ = 9 \), angle at \( K \) is \( 24^\circ \), so the side \( KJ \) is adjacent to angle \( K \), and \( IK \) is the hypotenuse? Wait, no, wait. Wait, \( IK \) is the hypotenuse? Wait, no, \( IK \): let's label the triangle. Points \( K \), \( J \), \( I \), with \( J \) being the right angle. So \( KJ \) is one leg (length 9), \( JI \) is another leg, and \( IK \) is the hypotenuse? Wait, no, \( IK \) is the hypotenuse? Wait, no, \( K \) to \( I \) is the hypotenuse. Wait, but angle at \( K \) is \( 24^\circ \), so the adjacent side to angle \( K \) is \( KJ = 9 \), and the hypotenuse is \( IK \). So \( \cos(24^\circ)=\frac{KJ}{IK} \), so \( IK = \frac{KJ}{\cos(24^\circ)} \). Let's calculate that. \( \cos(24^\circ) \approx 0.9135 \), so \( 9 / 0.9135 \approx 9.85 \), which rounds to 9.9? Wait, no, wait, maybe I mixed up the sides. Wait, maybe \( KJ \) is adjacent, and \( IK \) is the hypotenuse. Wait, but let's check again. Alternatively, maybe \( \sin(24^\circ) \)? Wait, no, angle at \( K \) is \( 24^\circ \), so the side opposite to angle \( K \) is \( JI \), and adjacent is \( KJ \). So hypotenuse is \( IK \). So \( \cos(24^\circ) = \frac{KJ}{IK} \), so \( IK = \frac{9}{\cos(24^\circ)} \approx \frac{9}{0.9135} \approx 9.85 \), which is approximately 9.9 when rounded to the nearest tenth? Wait, no, 9.85 rounded to the nearest tenth is 9.9? Wait, 9.85: the tenths place is 8, the hundredths is 5, so we round up the tenths place: 9.9. Wait, but let's use a calculator for more precision. \( \cos(24^\circ) \): let's use a calculator. \( \cos(24^\circ) \approx 0.9135454576 \). So \( 9 / 0.9135454576 \approx 9.851 \), which is 9.9 when rounded to the nearest tenth. Wait, but maybe I made a mistake in the trigonometric ratio. Wait, maybe \( KJ \) is adjacent, and \( IK \) is the hypotenuse. Alternatively, maybe \( \tan(24^\circ) \)? No, because \( \tan \) is opposite over adjacent. Wait, no, the problem is to find \( IK \). Let's re-express the triangle. Right triangle \( KJI \), right-angled at \( J \). So sides: \( KJ = 9 \) (horizontal leg), \( JI \) (vertical leg), \( IK \) (hypotenuse). Angle at \( K \) is \( 24^\circ \), so angle between \( KJ \) and \( IK \) is \( 24^\circ \). Therefore, \( \cos(24^\circ) = \frac{KJ}{IK} \), so \( IK = \frac{KJ}{\cos(24^\circ)} \). So that's correct. So calculating that, we get approximately 9.9. Wait, but let's check with another approach. Alternatively, maybe \( KJ \) is adjacent, and \( IK \) is the hypotenuse. So yes, the calculation is correct. So \( IK \approx 9.9 \). Wait, but wait, maybe I had the adjacent and hypotenuse wrong. Wait, no, in a right triangle, cosine of an angle is adjacent over hypotenuse. So angle at \( K \), adjacent side is \( KJ \), hypotenuse is \( IK \). So that's correct. So the calculation is \( IK = 9 / \cos(24^\circ) \approx 9 / 0.9135 \approx 9.85 \), which is 9.9 when rounded to the nearest tenth. Wait, but let's check with a calculator. Let me compute \( 9 / \cos(24^\circ) \). \( \cos(24^\circ) \approx 0.913545 \), so 9 divided by 0.913545 is approximately 9.851, which is 9.9 when rounded to the nearest tenth. Yes.

Answer:

\( 9.9 \)