Question

solve each triangle. (means to find all unknown sides rounded to 1 decimal place and angles to the nearest degree)
triangle with vertices j, k, l: side jk = 24, side jl = 25, angle at l is 58°

Explanation:

Step1: Use the Law of Cosines to find side \( KL \)

The Law of Cosines states that for a triangle with sides \( a \), \( b \), \( c \) and opposite angles \( A \), \( B \), \( C \) respectively, \( c^2 = a^2 + b^2 - 2ab\cos(C) \). In triangle \( JKL \), we know \( JL = 25 \), \( JK = 24 \), and \( \angle L = 58^\circ \). Let \( KL = x \), \( JL = c = 25 \), \( JK = b = 24 \), \( \angle L = C = 58^\circ \), and we want to find \( x = a \). Wait, actually, let's label the triangle properly: let \( J \), \( L \), \( K \) be the vertices, with \( JL = 25 \), \( JK = 24 \), \( \angle L = 58^\circ \). So side opposite \( \angle K \) is \( JL = 25 \), side opposite \( \angle J \) is \( KL \), and side opposite \( \angle L \) is \( JK = 24 \). Wait, maybe better to use Law of Cosines to find \( KL \) first. Wait, no, let's define: let \( JL = c = 25 \), \( JK = b = 24 \), \( \angle L = 58^\circ \), and we need to find \( KL = a \), \( \angle J \), \( \angle K \).

Wait, actually, the Law of Cosines for side \( JK \): \( JK^2 = JL^2 + KL^2 - 2 \cdot JL \cdot KL \cdot \cos(\angle L) \). Wait, \( JK = 24 \), \( JL = 25 \), \( \angle L = 58^\circ \), so:

\( 24^2 = 25^2 + KL^2 - 2 \cdot 25 \cdot KL \cdot \cos(58^\circ) \)

\( 576 = 625 + KL^2 - 50 \cdot KL \cdot \cos(58^\circ) \)

\( KL^2 - 50 \cos(58^\circ) \cdot KL + 625 - 576 = 0 \)

\( KL^2 - 50 \cos(58^\circ) \cdot KL + 49 = 0 \)

First, calculate \( \cos(58^\circ) \approx 0.5299 \)

So:

\( KL^2 - 50 \cdot 0.5299 \cdot KL + 49 = 0 \)

\( KL^2 - 26.495 \cdot KL + 49 = 0 \)

Using quadratic formula: \( KL = \frac{26.495 \pm \sqrt{(26.495)^2 - 4 \cdot 1 \cdot 49}}{2 \cdot 1} \)

Calculate discriminant: \( (26.495)^2 - 196 \approx 702.0 - 196 = 506.0 \) (wait, 26.495 squared is approx 26.5^2 = 702.25, so 702.25 - 196 = 506.25, square root of 506.25 is 22.5)

So \( KL = \frac{26.495 \pm 22.5}{2} \)

Two solutions:

\( KL = \frac{26.495 + 22.5}{2} \approx \frac{48.995}{2} \approx 24.5 \)

\( KL = \frac{26.495 - 22.5}{2} \approx \frac{3.995}{2} \approx 2.0 \)

But a side length of 2.0 would make the triangle impossible (since 2 + 24 = 26 > 25, but let's check angles). Wait, maybe I labeled the triangle wrong. Maybe \( \angle L \) is between \( JL \) and \( KL \), so sides \( JL = 25 \), \( KL = x \), and \( JK = 24 \) is opposite \( \angle L \). Wait, no, the Law of Sines might be better here. Let's use Law of Sines: \( \frac{JK}{\sin(\angle L)} = \frac{JL}{\sin(\angle K)} = \frac{KL}{\sin(\angle J)} \)

So \( \frac{24}{\sin(58^\circ)} = \frac{25}{\sin(\angle K)} \)

\( \sin(\angle K) = \frac{25 \cdot \sin(58^\circ)}{24} \)

\( \sin(58^\circ) \approx 0.8480 \)

\( \sin(\angle K) \approx \frac{25 \cdot 0.8480}{24} \approx \frac{21.2}{24} \approx 0.8833 \)

So \( \angle K \approx \arcsin(0.8833) \approx 62^\circ \) or \( 180^\circ - 62^\circ = 118^\circ \)

Now, check the sum of angles. If \( \angle K = 62^\circ \), then \( \angle J = 180^\circ - 58^\circ - 62^\circ = 60^\circ \)

If \( \angle K = 118^\circ \), then \( \angle J = 180^\circ - 58^\circ - 118^\circ = 4^\circ \), which seems too small, but let's check the side lengths.

First, take \( \angle K = 62^\circ \), \( \angle J = 60^\circ \)

Now, find \( KL \) using Law of Sines: \( \frac{KL}{\sin(\angle J)} = \frac{JK}{\sin(\angle L)} \)

\( KL = \frac{24 \cdot \sin(60^\circ)}{\sin(58^\circ)} \)

\( \sin(60^\circ) \approx 0.8660 \), \( \sin(58^\circ) \approx 0.8480 \)

\( KL \approx \frac{24 \cdot 0.8660}{0.8480} \approx \frac{20.784}{0.8480} \approx 24.5 \)

Now, check with Law of Cosines: \( KL^2 = 25^2 + 24^2 - 2 \cdot 25 \cdot 24 \cdot \cos(60^\circ) \)

\( \cos(60^\circ) = 0.5 \)

\( KL^2 = 625 + 576 - 2 \cdot 25 \cdot 24 \cdot 0.5 = 1201 - 600 = 601 \)

\( KL \approx \sqrt{601} \approx 24.5 \), which matches.

Now, check the other case: \( \angle K = 118^\circ \), \( \angle J = 4^\circ \)

\( KL = \frac{24 \cdot \sin(4^\circ)}{\sin(58^\circ)} \)

\( \sin(4^\circ) \approx 0.0698 \)

\( KL \approx \frac{24 \cdot 0.0698}{0.8480} \approx \frac{1.675}{0.8480} \approx 2.0 \)

Now, check with Law of Cosines: \( KL^2 = 25^2 + 24^2 - 2 \cdot 25 \cdot 24 \cdot \cos(4^\circ) \)

\( \cos(4^\circ) \approx 0.9976 \)

\( KL^2 = 625 + 576 - 2 \cdot 25 \cdot 24 \cdot 0.9976 = 1201 - 1197.12 = 3.88 \)

\( KL \approx \sqrt{3.88} \approx 1.97 \approx 2.0 \), which also matches. But which one is correct?

Looking at the triangle, \( JK = 24 \), \( JL = 25 \), so \( JL \) is slightly longer than \( JK \), so \( \angle K \) should be slightly larger than \( \angle L = 58^\circ \), so \( 62^\circ \) is more reasonable than \( 118^\circ \) (since \( 118^\circ \) would make \( \angle K \) much larger, and \( \angle J \) very small). So we take \( \angle K \approx 62^\circ \), \( \angle J \approx 60^\circ \), \( KL \approx 24.5 \)

Step2: Verify the angles sum to 180°

\( 58^\circ + 62^\circ + 60^\circ = 180^\circ \), which is correct.

Answer:

• Side \( KL \approx 24.5 \)
• Angle \( \angle J \approx 60^\circ \)
• Angle \( \angle K \approx 62^\circ \)