Question

1. in each diagram, solve for the indicated sides and angles

a)
diagram: a quadrilateral with two right angles, a side labeled 6.8 cm, angles 54° and 32°, and sides labeled x and y

Explanation:

Step1: Solve for \( y \) using cosine in the left triangle

In the left right - triangle, we know the adjacent side to the \( 32^{\circ} \) angle is \( 6.8 \) cm, and \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). So, \( \cos(32^{\circ})=\frac{6.8}{y} \), then \( y = \frac{6.8}{\cos(32^{\circ})} \). Calculate \( \cos(32^{\circ})\approx0.8480 \), so \( y\approx\frac{6.8}{0.8480}\approx7.90 \) cm.

Step2: Solve for \( x \) using sine in the right triangle or cosine in the left - right relation

We can also use the right - triangle on the right. We know the hypotenuse \( y\approx7.90 \) cm, and we can use \( \sin(54^{\circ})=\frac{x}{y} \) (since in the right - triangle on the right, \( x \) is the opposite side to the \( 54^{\circ} \) angle). So \( x = y\sin(54^{\circ}) \). We know \( \sin(54^{\circ})\approx0.8090 \), and \( y\approx7.90 \) cm. Then \( x\approx7.90\times0.8090\approx6.40 \) cm. Alternatively, we can use the left - triangle: \( \tan(32^{\circ})=\frac{\text{opposite}}{\text{adjacent}} \), but the opposite side in the left - triangle is the same as \( x \). \( \tan(32^{\circ})=\frac{x}{6.8} \), \( \tan(32^{\circ})\approx0.6249 \), so \( x = 6.8\times0.6249\approx4.25 \)? Wait, there is a mistake here. Wait, the two right - triangles share the hypotenuse \( y \). The left - triangle has angles \( 90^{\circ}, 32^{\circ} \), so the third angle is \( 58^{\circ} \)? No, wait the right - triangle on the right has a right angle and a \( 54^{\circ} \) angle, so the third angle is \( 36^{\circ} \)? Wait, no, the two triangles are connected by the common hypotenuse \( y \). Let's re - examine.

In the left right - triangle: angle \( = 32^{\circ} \), adjacent side \( = 6.8 \) cm, hypotenuse \( = y \), so \( \cos(32^{\circ})=\frac{6.8}{y}\Rightarrow y=\frac{6.8}{\cos(32^{\circ})}\approx\frac{6.8}{0.8480}\approx7.90 \) cm (correct).

In the right right - triangle: angle \( = 54^{\circ} \), hypotenuse \( = y\approx7.90 \) cm, opposite side \( = x \), so \( \sin(54^{\circ})=\frac{x}{y}\Rightarrow x = y\sin(54^{\circ})\approx7.90\times0.8090\approx6.40 \) cm. Or we can use the left - triangle: \( \sin(32^{\circ})=\frac{\text{opposite}}{y} \), the opposite side in the left - triangle is equal to \( x \) (since the two triangles are part of the same quadrilateral with two right angles, so the side \( x \) is the opposite side of the \( 32^{\circ} \) angle in the left - triangle). Wait, \( \sin(32^{\circ})=\frac{x}{y} \), so \( x = y\sin(32^{\circ})\approx7.90\times0.5299\approx4.19 \) cm. There is a contradiction here. Wait, the error is in the angle correspondence. Let's look at the diagram again. The left - triangle has a right angle, one angle \( 32^{\circ} \), so the non - right angle is \( 32^{\circ} \), and the right - triangle on the right has a right angle and one angle \( 54^{\circ} \), so the non - right angle is \( 54^{\circ} \). Since the two triangles share the hypotenuse \( y \), the sum of the non - right angles of the two triangles should be \( 32^{\circ}+54^{\circ}=86^{\circ} eq90^{\circ} \), which means my initial assumption is wrong. Wait, no, the quadrilateral has two right angles (the two small squares), so the two triangles are right - triangles with a common hypotenuse \( y \), and the angles at the top vertex: one is \( 54^{\circ} \), the other is such that \( 32^{\circ}+54^{\circ}+\text{other angles}=90^{\circ}+90^{\circ} \)? No, the sum of the interior angles of a quadrilateral is \( 360^{\circ} \), with two right angles (\( 90^{\circ}\times2 = 180^{\circ} \)), so the sum of the other two angles (the \( 32^{\circ} \) and \( 54^{\circ} \) related angles) should be \( 180^{\circ} \). Wait, \( 32^{\circ}+54^{\circ}=86^{\circ} eq180^{\circ} \), so there is a misinterpretation.

Wait, maybe the left - triangle has angle \( 32^{\circ} \) at the bottom, right angle on the left, so the angle at the top of the left - triangle is \( 58^{\circ} \), and the right - triangle on the right has a right angle on the right, angle \( 54^{\circ} \) at the top, so the angle at the bottom of the right - triangle is \( 36^{\circ} \). But the two bottom angles ( \( 32^{\circ} \) and the bottom angle of the right - triangle) should add up to something? No, the correct way is:

In the left right - triangle: legs are \( 6.8 \) cm (adjacent to \( 32^{\circ} \)) and \( x \) (opposite to \( 32^{\circ} \)), hypotenuse \( y \). So \( \cos(32^{\circ})=\frac{6.8}{y} \), \( \sin(32^{\circ})=\frac{x}{y} \).

In the right right - triangle: legs are \( x \) (adjacent to \( 54^{\circ} \)) and the other leg (let's say \( z \)), hypotenuse \( y \). So \( \cos(54^{\circ})=\frac{x}{y} \), \( \sin(54^{\circ})=\frac{z}{y} \).

Ah! Here is the key: \( \sin(32^{\circ})=\cos(54^{\circ}) \) because \( 32^{\circ}+54^{\circ}=86^{\circ} eq90^{\circ} \)? No, \( \sin\theta=\cos(90^{\circ}-\theta) \). So \( \sin(32^{\circ})=\cos(58^{\circ}) \), \( \cos(54^{\circ})=\sin(36^{\circ}) \). So my previous angle correspondence was wrong.

Wait, let's start over. Let's denote the common hypotenuse as \( y \).

In the left right - triangle (right angle on the left, vertical side \( 6.8 \) cm, horizontal side \( x \), hypotenuse \( y \)): angle at the bottom is \( 32^{\circ} \), so \( \cos(32^{\circ})=\frac{6.8}{y} \) (adjacent over hypotenuse), \( \sin(32^{\circ})=\frac{x}{y} \) (opposite over hypotenuse).

In the right right - triangle (right angle on the right, horizontal side \( x \), vertical side \( z \), hypotenuse \( y \)): angle at the top is \( 54^{\circ} \), so \( \cos(54^{\circ})=\frac{x}{y} \) (adjacent over hypotenuse), \( \sin(54^{\circ})=\frac{z}{y} \) (opposite over hypotenuse).

So we have \( \sin(32^{\circ})=\cos(54^{\circ}) \)? Let's calculate \( \sin(32^{\circ})\approx0.5299 \), \( \cos(54^{\circ})\approx0.5878 \). Not equal. So there is a mistake in the angle labels. Wait, maybe the angle at the top of the left - triangle is \( 54^{\circ} \) and the angle at the bottom of the right - triangle is \( 32^{\circ} \). Let's re - assign:

Left right - triangle: right angle on the left, angle at the top \( 54^{\circ} \), so angle at the bottom is \( 36^{\circ} \), adjacent side to \( 54^{\circ} \) is \( x \), opposite side is \( 6.8 \) cm, hypotenuse \( y \). Then \( \sin(54^{\circ})=\frac{6.8}{y} \), \( \cos(54^{\circ})=\frac{x}{y} \).

Right right - triangle: right angle on the right, angle at the bottom \( 32^{\circ} \), so angle at the top is \( 58^{\circ} \), adjacent side to \( 32^{\circ} \) is \( 6.8 \) cm, opposite side is \( x \), hypotenuse \( y \). Then \( \cos(32^{\circ})=\frac{6.8}{y} \), \( \sin(32^{\circ})=\frac{x}{y} \).

Now, \( \sin(54^{\circ})\approx0.8090 \), \( \cos(32^{\circ})\approx0.8480 \). These are not equal, so the diagram's angle labels must have the two non - right angles (the \( 32^{\circ} \) and \( 54^{\circ} \)) such that \( 32^{\circ}+54^{\circ}=90^{\circ} \)? But \( 32 + 54=86 eq90 \). Wait, maybe it's a typo, and one of the angles is \( 36^{\circ} \) (since \( 36 + 54 = 90 \)) or \( 58^{\circ} \) (since \( 32+58 = 90 \)). Assuming that there is a mistake and the angles are complementary (sum to \( 90^{\circ} \)), let's assume the \( 32^{\circ} \) and \( 58^{\circ} \) (instead of \( 54^{\circ} \)) or \( 54^{\circ} \) and \( 36^{\circ} \) (instead of \( 32^{\circ} \)).

Assuming that the two non - right angles of the two right - triangles (sharing the hypotenuse) are complementary (sum to \( 90^{\circ} \)), so if one is \( 32^{\circ} \), the other is \( 58^{\circ} \), but the diagram shows \( 54^{\circ} \). Alternatively, maybe the diagram has the left - triangle with angle \( 32^{\circ} \), right - triangle with angle \( 54^{\circ} \), and we can solve for \( y \) and \( x \) as follows:

For \( y \): In the left right - triangle, \( \cos(32^{\circ})=\frac{6.8}{y}\Rightarrow y=\frac{6.8}{\cos(32^{\circ})}\approx\frac{6.8}{0.8480}\approx7.90 \) cm.

For \( x \): In the right right - triangle, \( \sin(54^{\circ})=\frac{x}{y}\Rightarrow x = y\sin(54^{\circ})\approx7.90\times0.8090\approx6.40 \) cm. Or in the left right - triangle, \( \tan(32^{\circ})=\frac{x}{6.8}\Rightarrow x = 6.8\times\tan(32^{\circ})\approx6.8\times0.6249\approx4.25 \) cm. There is a conflict because the angles are not complementary.

Wait, maybe the diagram is a quadrilateral with two right angles, and the diagonal divides it into two right - triangles. So the sum of the angles of the quadrilateral is \( 360^{\circ} \), two right angles (\( 90\times2 = 180 \)), so the sum of the other two angles (the ones with \( 32^{\circ} \) and \( 54^{\circ} \)) is \( 180^{\circ} \). So \( 32 + 54+\text{the angles at the diagonal}=180 \)? No, the two triangles each have a right angle, so the sum of all angles in the two triangles is \( 180\times2 = 360 \), which matches the quadrilateral's angle sum. So the angles at the diagonal: in the left - triangle, angles are \( 90^{\circ}, 32^{\circ}, \alpha \); in the right - triangle, angles are \( 90^{\circ}, 54^{\circ}, \beta \), and \( \alpha+\beta = 180^{\circ}-(90 + 32)-(90 + 54)+180^{\circ} \)? No, this is getting too complicated. Let's use the first approach with the given angles:

1. Solve for \( y \):
In the left right - triangle, \( \cos(32^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{6.8}{y} \)
\( y=\frac{6.8}{\cos(32^{\circ})} \)
\( \cos(32^{\circ})\approx0.8480 \)
\( y\approx\frac{6.8}{0.8480}\approx7.90 \) cm

2. Solve for \( x \):
In the right right - triangle, \( \sin(54^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{y} \)
\( x = y\sin(54^{\circ}) \)
\( \sin(54^{\circ})\approx0.8090 \)
\( x\approx7.90\times0.8090\approx6.40 \) cm

Or using the left - triangle: \( \tan(32^{\circ})=\frac{x}{6.8} \)
\( x = 6.8\times\tan(32^{\circ})\approx6.8\times0.6249\approx4.25 \) cm

There is a mistake in the angle relationship. The correct way is that if the two triangles are right - triangles with a common hypotenuse, then the angles opposite to the common side (the legs) should satisfy \( \sin(\theta_1)=\cos(\theta_2) \) where \( \theta_1+\theta_2 = 90^{\circ} \). Since \( 32^{\circ}+54^{\circ}=86^{\circ} eq90^{\circ} \), there is an error in the problem's angle labeling. But assuming the problem has a typo and the angles are complementary (e.g., \( 32^{\circ} \) and \( 58^{\circ} \) or \( 54^{\circ} \) and \( 36^{\circ} \)), but with the given angles, we can proceed with the calculation as:

If we take \( y=\frac{6.8}{\cos(32^{\circ})}\approx7.90 \) cm and \( x = y\sin(54^{\circ})\approx6.40 \) cm (using the right - triangle on the right) or \( x = 6.8\tan(32^{\circ})\approx4.25 \) cm (using the left - triangle). The discrepancy is due to the non - complementary angles.

Answer:

\( y\approx7.90 \) cm, \( x\approx6.40 \) cm (or \( x\approx4.25 \) cm depending on the angle interpretation)