Question

use one of the triangles to approximate the ratio $\frac{kl}{jl}$.
choose 1 answer:
a 0.64
b 0.77
c 0.83
d 1.9

Explanation:

Step1: Identify the triangle's angle and sides

We have a right - triangle \( \triangle JKL \) with \( \angle L = 90^{\circ} \) and \( \angle J=40^{\circ} \). We can use a similar right - triangle from the given set of triangles. Let's consider the first triangle (triangle 1) with legs of length \( 8.4 \) and \( 8.4 \)? Wait, no, let's look at the angles. The ratio \( \frac{KL}{JL} \) is \( \tan(\angle J) \) or \( \cot(\angle K) \)? Wait, in right - triangle \( JKL \), \( \tan(\angle J)=\frac{KL}{JL} \), since \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \), where \( \theta = \angle J = 40^{\circ} \), opposite side to \( \angle J \) is \( KL \) and adjacent side is \( JL \).

Alternatively, we can use the given triangles. Let's take the triangle with sides \( 8.4 \), \( 7.5 \) and hypotenuse \( 10 \)? Wait, the third triangle has sides \( 8.4 \), \( 7.5 \) and hypotenuse \( 10 \)? Wait, no, the third triangle has a right angle, one leg \( 8.4 \), one leg \( 7.5 \) and hypotenuse \( 10 \)? Wait, let's check the angles. The triangle with angle \( 40^{\circ} \) (wait, no, the first triangle has a \( 50^{\circ} \) angle? Wait, the first triangle: right - triangle with legs \( 8.4 \) and \( 8.4 \)? No, the first triangle has a right angle, one leg \( 8.4 \), one leg \( 8.4 \)? Wait, no, the first triangle: right - triangle, one angle \( 50^{\circ} \), legs \( 8.4 \) and \( 8.4 \)? Wait, maybe we can use the triangle with angle \( 40^{\circ} \) related. Wait, let's calculate \( \frac{8.4}{10} = 0.84 \)? No, wait, let's take the triangle with sides \( 7.5 \) and \( 8.4 \). Wait, the ratio \( \frac{7.5}{8.4}\approx0.89 \)? No, wait, maybe the triangle with angle \( 40^{\circ} \) - let's use the tangent of \( 40^{\circ} \). \( \tan(40^{\circ})\approx0.8391\approx0.84 \)? Wait, no, the options are \( 0.64 \), \( 0.77 \), \( 0.83 \), \( 1.2 \). Wait, maybe we made a mistake. Wait, \( \frac{KL}{JL} \): if we consider the triangle with sides \( 8.4 \) and \( 10 \)? No, let's take the triangle with angle \( 40^{\circ} \). Wait, another approach: in the triangle with legs \( 7.5 \) and \( 8.4 \), \( \frac{7.5}{8.4}\approx0.89 \), no. Wait, the first triangle: right - triangle, one leg \( 8.4 \), one leg \( 8.4 \)? No, the first triangle has a right angle, one angle \( 50^{\circ} \), so the other angle is \( 40^{\circ} \). So \( \tan(40^{\circ})=\frac{\text{opposite}}{\text{adjacent}} \). If the adjacent side to \( 40^{\circ} \) is \( 8.4 \) and opposite side is \( 7.5 \)? No, wait, let's calculate \( \frac{7.5}{8.4}\approx0.89 \), no. Wait, the ratio \( \frac{KL}{JL} \): let's look at the triangle with sides \( 8.4 \) and \( 10 \). Wait, \( \frac{8.4}{10} = 0.84 \), but the option is \( 0.83 \) which is close. Wait, maybe the triangle with angle \( 40^{\circ} \), \( \tan(40^{\circ})\approx0.839\approx0.83 \).

Wait, let's use the triangle with sides \( 7.5 \) and \( 8.4 \). Wait, no, let's take the triangle where we can find the ratio. Let's assume that the triangle with angle \( 40^{\circ} \) has \( KL \) and \( JL \) such that \( \frac{KL}{JL}=\tan(40^{\circ})\approx0.839\approx0.83 \).

Step2: Calculate the ratio

We know that in a right - triangle, for an angle \( \theta \), \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For \( \angle J = 40^{\circ} \), \( \tan(40^{\circ})\approx0.839\approx0.83 \).

Answer:

\( 0.83 \) (Option C)