Question

1. the vertex angle of an isosceles triangle is 64° and the length of its base is 24 in. find the length of one of its congruent sides.

Explanation:

Step1: Divide the isosceles triangle

Draw the altitude from the vertex angle to the base, which bisects the base and the vertex angle. So the base is split into two segments of \( \frac{24}{2}=12 \) inches, and the vertex angle is split into two angles of \( \frac{64^{\circ}}{2} = 32^{\circ}\).

Step2: Use trigonometric ratio

Let the length of the congruent side be \( x \). In the right triangle formed (half of the isosceles triangle), we have \( \cos(32^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{x} \).

Step3: Solve for \( x \)

Rearrange the formula to \( x = \frac{12}{\cos(32^{\circ})} \). Calculate \( \cos(32^{\circ})\approx0.8480 \), so \( x=\frac{12}{0.8480}\approx14.15 \) inches. Wait, no, wait, actually, the angle at the base? Wait, no, when we draw the altitude, the angle at the vertex of the right triangle is \( 32^{\circ} \), the adjacent side is 12, hypotenuse is \( x \) (the congruent side). Wait, no, maybe I mixed up. Wait, the isosceles triangle has two equal sides (legs) and a base. The vertex angle is between the two equal sides. So when we drop the altitude, it splits the base into two equal parts (12 each) and splits the vertex angle into two equal angles (32 each). So in the right triangle, the angle at the vertex (of the right triangle) is 32 degrees, the opposite side? No, wait, the adjacent side to the 32 - degree angle is 12? Wait, no, the angle at the base: wait, no, let's re - define. Let's call the isosceles triangle \( \triangle ABC \), with \( AB = AC=x \) (congruent sides), \( BC = 24 \) (base), and \( \angle BAC = 64^{\circ} \). Draw altitude \( AD \) from \( A \) to \( BC \), so \( D \) is the mid - point of \( BC \), \( BD = 12 \), \( \angle BAD=\angle CAD = 32^{\circ} \), and \( \triangle ABD \) is a right triangle with \( \angle ADB = 90^{\circ} \), \( \angle BAD = 32^{\circ} \), \( BD = 12 \), and \( AB=x \). In \( \triangle ABD \), \( \sin(32^{\circ})=\frac{BD}{AB} \)? No, \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \), \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \). Wait, \( \angle BAD = 32^{\circ} \), the side opposite to \( 32^{\circ} \) is \( BD = 12 \), and the hypotenuse is \( AB=x \). So \( \sin(32^{\circ})=\frac{12}{x} \), so \( x=\frac{12}{\sin(32^{\circ})} \). \( \sin(32^{\circ})\approx0.5299 \), so \( x=\frac{12}{0.5299}\approx22.65 \) inches. Ah, I made a mistake earlier. The angle at \( A \) in the right triangle is \( 32^{\circ} \), the side opposite to it is \( BD = 12 \), so we use sine. Let's re - derive:

In \( \triangle ABD \), \( \angle BAD = 32^{\circ} \), \( \angle ADB = 90^{\circ} \), \( BD = 12 \), \( AB=x \).

\( \sin(\angle BAD)=\frac{BD}{AB} \)

So \( \sin(32^{\circ})=\frac{12}{x} \)

\( x=\frac{12}{\sin(32^{\circ})} \)

\( \sin(32^{\circ})\approx0.5299 \)

\( x=\frac{12}{0.5299}\approx22.65 \) inches. Wait, let's check with cosine. The angle at \( B \): in the isosceles triangle, the base angles are \( \frac{180 - 64}{2}=58^{\circ} \). So in \( \triangle ABD \), \( \angle ABD = 58^{\circ} \), \( \angle ADB = 90^{\circ} \), \( BD = 12 \), \( AB=x \). Then \( \cos(58^{\circ})=\frac{BD}{AB} \), \( \cos(58^{\circ})\approx0.5299 \), so \( x=\frac{12}{\cos(58^{\circ})}\approx\frac{12}{0.5299}\approx22.65 \) inches. Yes, that's correct. Because \( \cos(58^{\circ})=\sin(32^{\circ}) \). So the length of the congruent side is approximately \( \frac{12}{\cos(58^{\circ})}\approx22.65 \) inches (or \( \frac{12}{\sin(32^{\circ})}\approx22.65 \) inches).

Wait, let's do it properly. The base angles of an isosceles triangle are equal. So base angle \( \beta=\frac{180 - \alpha}{2} \), where \( \alpha = 64^{\circ} \), so \( \beta=\frac{180 - 64}{2}=58^{\circ} \). Then, using the Law of Sines: \( \frac{a}{\sin\alpha}=\frac{b}{\sin\beta} \), where \( a = 24 \) (base), \( b=x \) (congruent side). So \( \frac{24}{\sin(64^{\circ})}=\frac{x}{\sin(58^{\circ})} \). Then \( x=\frac{24\sin(58^{\circ})}{\sin(64^{\circ})} \). Calculate \( \sin(64^{\circ})\approx0.8988 \), \( \sin(58^{\circ})\approx0.8480 \). So \( x=\frac{24\times0.8480}{0.8988}=\frac{20.352}{0.8988}\approx22.64 \) inches. Yes, that's consistent.

Answer:

The length of one of the congruent sides is approximately \( \boldsymbol{22.6} \) inches (or more precisely \( \approx22.64 \) inches).