Question

1. in right triangle jkl in the diagram below, kl=7, jk=24, jl=25, and \\(\angle k = 90^\circ\\).

diagram of right triangle jkl with right angle at k, jk=24, kl=7, jl=25

which statement is not true?

1. \\(\tan l = \frac{24}{7}\\)
2. \\(\cos l = \frac{24}{25}\\)
3. \\(\tan j = \frac{7}{24}\\)
4. \\(\sin j = \frac{7}{25}\\)

1. in \\(\triangle abc\\), \\(m\angle c = 90\\). if \\(ab = 5\\) and \\(ac = 4\\), which statement is not true?
2. \\(\cos a = \frac{4}{5}\\)
3. \\(\tan a = \frac{3}{4}\\)
4. \\(\sin b = \frac{4}{5}\\)
5. \\(\tan b = \frac{5}{3}\\) (with option 4 circled)

Explanation:

Question 19 (Right Triangle \(JKL\))

Step 1: Recall Trigonometric Ratios

In a right triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\triangle JKL\) with \(\angle K = 90^\circ\), sides: \(KL = 7\) (opposite \(\angle J\), adjacent \(\angle L\)), \(JK = 24\) (adjacent \(\angle J\), opposite \(\angle L\)), \(JL = 25\) (hypotenuse).

Step 2: Analyze Option 1 (\(\tan L\))

\(\tan L=\frac{\text{opposite to }L}{\text{adjacent to }L}=\frac{JK}{KL}=\frac{24}{7}\). True.

Step 3: Analyze Option 2 (\(\cos L\))

\(\cos L=\frac{\text{adjacent to }L}{\text{hypotenuse}}=\frac{KL}{JL}=\frac{7}{25}
eq\frac{24}{25}\). False (but we check others to confirm).

Step 4: Analyze Option 3 (\(\tan J\))

\(\tan J=\frac{\text{opposite to }J}{\text{adjacent to }J}=\frac{KL}{JK}=\frac{7}{24}\). True.

Step 5: Analyze Option 4 (\(\sin J\))

\(\sin J=\frac{\text{opposite to }J}{\text{hypotenuse}}=\frac{KL}{JL}=\frac{7}{25}\). True.

Step 1: Find \(BC\) using Pythagoras

In \(\triangle ABC\), \(\angle C = 90^\circ\), \(AB = 5\) (hypotenuse), \(AC = 4\). So \(BC=\sqrt{AB^2 - AC^2}=\sqrt{25 - 16}=\sqrt{9}=3\).

Step 2: Recall Trigonometric Ratios

For \(\angle A\): adjacent \(= AC = 4\), opposite \(= BC = 3\), hypotenuse \(= AB = 5\). For \(\angle B\): adjacent \(= BC = 3\), opposite \(= AC = 4\), hypotenuse \(= AB = 5\).

Step 3: Analyze Option 1 (\(\cos A\))

\(\cos A=\frac{\text{adjacent to }A}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{4}{5}\). True.

Step 4: Analyze Option 2 (\(\tan A\))

\(\tan A=\frac{\text{opposite to }A}{\text{adjacent to }A}=\frac{BC}{AC}=\frac{3}{4}\). True.

Step 5: Analyze Option 3 (\(\sin B\))

\(\sin B=\frac{\text{opposite to }B}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{4}{5}\). True.

Step 6: Analyze Option 4 (\(\tan B\))

\(\tan B=\frac{\text{opposite to }B}{\text{adjacent to }B}=\frac{AC}{BC}=\frac{4}{3}
eq\frac{5}{3}\). False.

Answer:

1. \(\cos L=\frac{24}{25}\)

Question 20 (Right Triangle \(ABC\))