Question

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

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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 \( x = a \) (opposite \( \angle J \))? Wait, no, actually, let's label correctly: Let \( J \), \( L \), \( K \) be the vertices. So sides: \( JL = 25 \) (between \( J \) and \( L \)), \( JK = 24 \) (between \( J \) and \( K \)), and we need to find \( KL \) (between \( K \) and \( L \)), and angles \( \angle J \) and \( \angle K \).

Using Law of Cosines for side \( KL \) (let's call it \( k \), opposite \( \angle J \); \( JL = l = 25 \), opposite \( \angle K \); \( JK = j = 24 \), opposite \( \angle L = 58^\circ \)). Wait, maybe better to use Law of Cosines for side \( KL \):

\( JK^2 = JL^2 + KL^2 - 2 \cdot JL \cdot KL \cdot \cos(\angle L) \)

Wait, no: \( JK \) is between \( J \) and \( K \), so the sides adjacent to \( \angle L \) are \( JL = 25 \) and \( KL = x \), and the side opposite \( \angle L \) is \( JK = 24 \). So Law of Cosines: \( JK^2 = JL^2 + KL^2 - 2 \cdot JL \cdot KL \cdot \cos(\angle L) \)

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

Calculate \( 24^2 = 576 \), \( 25^2 = 625 \), \( \cos(58^\circ) \approx 0.5299 \)

So \( 576 = 625 + x^2 - 2 \cdot 25 \cdot x \cdot 0.5299 \)

Simplify: \( 576 = 625 + x^2 - 26.495x \)

Rearrange: \( x^2 - 26.495x + 625 - 576 = 0 \)

\( x^2 - 26.495x + 49 = 0 \)

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

Calculate discriminant: \( (26.495)^2 - 196 \approx 702.0 - 196 = 506.0 \) (approx, more accurately: \( 26.495^2 = (26 + 0.495)^2 = 26^2 + 2 \cdot 26 \cdot 0.495 + 0.495^2 = 676 + 25.74 + 0.245 = 701.985 \), so discriminant \( 701.985 - 196 = 505.985 \), square root of 505.985 ≈ 22.5

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

Two solutions: \( x = \frac{26.495 + 22.5}{2} \approx \frac{48.995}{2} \approx 24.5 \), or \( x = \frac{26.495 - 22.5}{2} \approx \frac{3.995}{2} \approx 2.0 \). But since \( JL = 25 \) and \( JK = 24 \), \( KL \) can't be 2.0 (too small), so \( KL \approx 24.5 \)

Step2: Use Law of Sines to find \( \angle J \)

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

We know \( KL \approx 24.5 \), \( \angle L = 58^\circ \), \( JK = 24 \)

So \( \sin(\angle J) = \frac{KL \cdot \sin(\angle L)}{JK} = \frac{24.5 \cdot \sin(58^\circ)}{24} \)

\( \sin(58^\circ) \approx 0.8480 \), so \( \sin(\angle J) \approx \frac{24.5 \cdot 0.8480}{24} \approx \frac{20.776}{24} \approx 0.8657 \)

So \( \angle J \approx \arcsin(0.8657) \approx 59.9^\circ \approx 60^\circ \)

Step3: Find \( \angle K \)

Since the sum of angles in a triangle is \( 180^\circ \), \( \angle K = 180^\circ - \angle J - \angle L \approx 180 - 60 - 58 = 62^\circ \) (wait, let's check with more accurate \( \angle J \))

Wait, \( \arcsin(0.8657) \) is actually about \( 60^\circ \) (since \( \sin(60^\circ) \approx 0.8660 \), which is very close). So \( \angle J \approx 60^\circ \), then \( \angle K = 180 - 60 - 58 = 62^\circ \)

Wait, let's verify with Law of Cosines for \( \angle J \):

\( \cos(\angle J) = \frac{JK^2 + JL^2 - KL^2}{2 \cdot JK \cdot JL} = \frac{24^2 + 25^2 - 24.5^2}{2 \cdot 24 \cdot 25} \)

Calculate numerator: \( 576 + 625 - 600.25 = 1201 - 600.25 = 600.75 \)

Denominator: \( 1200 \)

So \( \cos(\angle J) = \frac{600.75}{1200} \approx 0.5006 \), so \( \angle J \approx \arccos(0.5006) \approx 59.9^\circ \approx 60^\circ \), which matches.

Then \( \angle K = 180 - 60 - 58 = 62^\circ \)

Answer:

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