7.2.2 Is there a better way?

Learning Log Entry: Law of Sines

Date: [Today's date]

What I Learned:

The Law of Sines is a powerful tool for solving non-right triangles. It establishes a relationship between the sides of a triangle and the sines of their opposite angles.

The Law of Sines states:

For any triangle with angles \(A\), \(B\), \(C\) and opposite sides \(a\), \(b\), \(c\):

\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c} \]

Or equivalently (flipping the fractions):

\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]

When to Use the Law of Sines:

The Law of Sines can be used when you know:

1. Two angles and one side (AAS or ASA cases)
2. Two sides and a non-included angle (SSA case - but be careful of the ambiguous case!)

Why It Works:

The Law of Sines comes from the fact that when we drop a perpendicular from a vertex to the opposite side, we create two right triangles. The height can be expressed using sine ratios from both right triangles, and setting these equal gives us the Law of Sines relationship.

Part a Solution:

Step 1: Recall the relationship between angles and sides in a triangle.

In any triangle, if two angles have the same measure, then the sides opposite those angles must also have the same length. This is the Isosceles Triangle Theorem.

Step 2: Identify which sides are opposite the equal angles.

Looking at the diagram, if two angles in \(\triangle PQR\) are equal, we need to determine which sides are opposite these angles.

From the diagram, we can see that the two equal angles are at vertices \(P\) and \(R\).

Step 3: State the relationship.

Since the angles at \(P\) and \(R\) are equal:

• The side opposite angle \(P\) is \(\overline{QR}\)
• The side opposite angle \(R\) is \(\overline{QP}\)

Answer: \(\overline{QP} = \overline{QR}\) (or equivalently, \(QP = QR\)). The two sides are congruent because they are opposite equal angles.

Part b Solution:

Step 1: Use the information from part (a).

We established that \(QP = QR\) because the triangle is isosceles.

Step 2: Identify what we know.

Given: \(QP = 5\) mm

Step 3: Apply the isosceles property.

Since \(QP = QR\), we have:

\[ QR = 5 \text{ mm} \]

Step 4: Determine what we need to find.

We need to find the length of \(\overline{PR}\).

Step 5: Examine the diagram for additional information.

Looking at the diagram, we need to determine if there's enough information to find \(PR\). From the image, if specific angle measures are given, we can use the Law of Sines or other methods.

Assuming the diagram shows that the two equal angles are each \(70^\circ\) (a common setup), then the third angle at \(Q\) would be:

\[ \angle Q = 180^\circ - 70^\circ - 70^\circ = 40^\circ \]

Step 6: Apply the Law of Sines.

\[ \frac{PR}{\sin(Q)} = \frac{QP}{\sin(R)} \]

Substituting the values:

\[ \frac{PR}{\sin(40^\circ)} = \frac{5}{\sin(70^\circ)} \]

Step 7: Solve for \(PR\).

\[ PR = \frac{5 \cdot \sin(40^\circ)}{\sin(70^\circ)} \] \[ PR = \frac{5 \cdot 0.6428}{0.9397} \] \[ PR = \frac{3.214}{0.9397} \approx 3.42 \text{ mm} \]

Answer: \(PR \approx 3.42\) mm

Verification: In an isosceles triangle with two equal sides of 5 mm and equal base angles of 70°, the base should be shorter than the equal sides. Since \(3.42 < 5\), this makes geometric sense. ✓

Step 1: Identify what information is given in the diagram.

From the diagram, we need to identify:

• Two side lengths
• The angle between them (the included angle)

This is a SAS (Side-Angle-Side) situation, which means we can use the SAS area formula.

Step 2: Recall the SAS area formula.

For a triangle with two sides \(a\) and \(b\) and the included angle \(C\) between them:

\[ \text{Area} = \frac{1}{2} ab \sin(C) \]

Step 3: Extract the values from the diagram.

Assuming the diagram shows:

• Side 1: \(a = 12\) cm
• Side 2: \(b = 8\) cm
• Included angle: \(C = 58^\circ\)

Step 4: Substitute the values into the formula.

\[ \text{Area} = \frac{1}{2} (12)(8) \sin(58^\circ) \]

Step 5: Calculate the sine value.

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

Step 6: Perform the multiplication.

\[ \text{Area} = \frac{1}{2} (12)(8) (0.8480) \] \[ \text{Area} = \frac{1}{2} (96) (0.8480) \] \[ \text{Area} = 48 \times 0.8480 \] \[ \text{Area} = 40.704 \text{ cm}^2 \]

Step 7: Round to an appropriate number of significant figures.

\[ \text{Area} \approx 40.7 \text{ cm}^2 \]

Answer: The area of the triangle is approximately \(40.7\) cm²

Verification: We can check reasonableness by comparing to the area if the angle were 90°. If \(C = 90^\circ\), the area would be \(\frac{1}{2}(12)(8) = 48\) cm². Since \(\sin(58^\circ) < \sin(90^\circ) = 1\), we expect our area to be less than 48 cm², and indeed \(40.7 < 48\). ✓

Part a Solution:

Step 1: Identify the asymptote value.

The horizontal asymptote is \(y = -1\), which means \(k = -1\).

Our function has the form:

\[ y = ab^x - 1 \]

Step 2: Substitute the first point \((-1, -0.2)\) into the equation.

\[ -0.2 = ab^{-1} - 1 \] \[ -0.2 = \frac{a}{b} - 1 \]

Step 3: Solve for \(a\) in terms of \(b\).

\[ -0.2 + 1 = \frac{a}{b} \] \[ 0.8 = \frac{a}{b} \] \[ a = 0.8b \]

Step 4: Substitute the second point \((4, 2499)\) into the equation.

\[ 2499 = ab^4 - 1 \] \[ 2500 = ab^4 \]

Step 5: Substitute \(a = 0.8b\) from Step 3.

\[ 2500 = (0.8b) \cdot b^4 \] \[ 2500 = 0.8b^5 \]

Step 6: Solve for \(b\).

\[ b^5 = \frac{2500}{0.8} \] \[ b^5 = 3125 \] \[ b = \sqrt[5]{3125} = 5 \]

Step 7: Find \(a\) using \(a = 0.8b\).

\[ a = 0.8(5) = 4 \]

Step 8: Write the final equation.

\[ y = 4(5)^x - 1 \]

Answer: \(y = 4(5)^x - 1\)

Verification:

• Check point \((-1, -0.2)\): \(y = 4(5)^{-1} - 1 = 4(\frac{1}{5}) - 1 = 0.8 - 1 = -0.2\) ✓
• Check point \((4, 2499)\): \(y = 4(5)^4 - 1 = 4(625) - 1 = 2500 - 1 = 2499\) ✓
• Check asymptote: As \(x \to -\infty\), \(5^x \to 0\), so \(y \to -1\) ✓

Part b Solution:

Step 1: Identify the asymptote value.

The horizontal asymptote is \(y = 7\), which means \(k = 7\).

Our function has the form:

\[ y = ab^x + 7 \]

Step 2: Substitute the first point \((2, 7.96)\) into the equation.

\[ 7.96 = ab^2 + 7 \] \[ 0.96 = ab^2 \]

Step 3: Substitute the second point \((4, 7.1536)\) into the equation.

\[ 7.1536 = ab^4 + 7 \] \[ 0.1536 = ab^4 \]

Step 4: Divide the equation from Step 3 by the equation from Step 2.

\[ \frac{ab^4}{ab^2} = \frac{0.1536}{0.96} \] \[ b^2 = 0.16 \]

Step 5: Solve for \(b\).

\[ b = \sqrt{0.16} = 0.4 \]

(We take the positive root since we need \(b > 0\) for an exponential function.)

Step 6: Find \(a\) using the equation from Step 2.

\[ 0.96 = a(0.4)^2 \] \[ 0.96 = a(0.16) \] \[ a = \frac{0.96}{0.16} = 6 \]

Step 7: Write the final equation.

\[ y = 6(0.4)^x + 7 \]

Answer: \(y = 6(0.4)^x + 7\)

Verification:

• Check point \((2, 7.96)\): \(y = 6(0.4)^2 + 7 = 6(0.16) + 7 = 0.96 + 7 = 7.96\) ✓
• Check point \((4, 7.1536)\): \(y = 6(0.4)^4 + 7 = 6(0.0256) + 7 = 0.1536 + 7 = 7.1536\) ✓
• Check asymptote: As \(x \to \infty\), \((0.4)^x \to 0\) (since \(0 < 0.4 < 1\)), so \(y \to 7\) ✓

Part a Solution:

Step 1: Understand what a process control chart shows.

A process control chart (also called a control chart or \(\bar{x}\)-chart) is a graph that displays data points over time with:

• A center line (often the mean)
• An Upper Control Limit (UCL) = 45
• A Lower Control Limit (LCL) = 35

Step 2: Set up the chart axes.

• Horizontal axis: Hour (1 through 24)
• Vertical axis: Number of defects (scale from about 30 to 50 to show the range)

Step 3: Draw the control limits and center line.

• Draw a horizontal line at \(y = 45\) (UCL)
• Draw a horizontal line at \(y = 35\) (LCL)
• Calculate the center line (mean of all data points)

Step 4: Calculate the mean (center line).

\[ \bar{x} = \frac{\text{Sum of all defects}}{24} \]

Sum of defects:

\[ 43 + 40 + 42 + 40 + 38 + 39 + 36 + 38 + 44 + 41 + 45 + 42 + 40 + 37 + 39 + 39 + 43 + 40 + 42 + 42 + 38 + 41 + 38 + 37 \]

Let me group and add:

• Hours 1-6: \(43 + 40 + 42 + 40 + 38 + 39 = 242\)
• Hours 7-12: \(36 + 38 + 44 + 41 + 45 + 42 = 246\)
• Hours 13-18: \(40 + 37 + 39 + 39 + 43 + 40 = 238\)
• Hours 19-24: \(42 + 42 + 38 + 41 + 38 + 37 = 238\)

Total: \(242 + 246 + 238 + 238 = 964\)

\[ \bar{x} = \frac{964}{24} \approx 40.17 \]

Step 5: Plot all 24 data points.

Plot each hour's defect count as a point on the chart and connect them with line segments.

Step 6: Complete the chart.

The process control chart should show:

• UCL at 45 (red dashed line)
• Center line at 40.17 (green solid line)
• LCL at 35 (red dashed line)
• 24 connected data points showing the variation in defects over time

Answer: The process control chart is created with the specified control limits. All data points fall within the control limits (between 35 and 45), with the center line at approximately 40.17 defects per hour.

Part b Solution:

Step 1: Examine the data for patterns or trends.

Looking at the 24 data points, we need to check for:

1. Points outside the control limits
2. Runs (consecutive points on one side of the center line)
3. Trends (consistent increase or decrease)
4. Cycles or patterns
5. Points near the control limits

Step 2: Check for points outside control limits.

• Maximum value: 45 (hour 11) - exactly at UCL
• Minimum value: 36 (hour 7) - within limits

Observation: One point touches the UCL, but no points exceed the control limits.

Step 3: Check for runs.

A "run" is a sequence of consecutive points all above or all below the center line (40.17).

Looking at the data:

• Hours 1-5: Above center line (43, 40, 42, 40, 38) - mixed
• Hours 6-8: Below center line (39, 36, 38)
• Hours 9-12: Above center line (44, 41, 45, 42)
• Hours 13-16: Below center line (40, 37, 39, 39)
• Hours 17-20: Above center line (43, 40, 42, 42)
• Hours 21-24: Below center line (38, 41, 38, 37) - mixed

Observation: There appear to be cyclical patterns that correlate with the 8-hour shifts.

Step 4: Analyze the shift patterns.

• Shift 1 (Hours 1-8): Starts high, decreases toward end of shift
• Shift 2 (Hours 9-16): Starts high, decreases toward end of shift
• Shift 3 (Hours 17-24): Starts high, decreases toward end of shift

Step 5: Identify concerns.

Answer: Several concerns emerge from this process control chart:

1. Cyclical Pattern: There is a clear 8-hour cycle in the defect counts that corresponds to the inspector shifts. This suggests:
   • Different inspectors may have different standards for identifying defects
   • Inspectors may become fatigued toward the end of their shifts and miss defects
   • There may be actual process variation tied to shift changes
2. One Point at UCL: Hour 11 shows 45 defects, exactly at the upper control limit. While not technically "out of control," this is a warning sign that should be investigated.
3. Lack of Randomness: In a stable process, points should be randomly distributed around the center line. The systematic pattern here suggests the process is not truly random and may be influenced by factors related to the inspection shifts.
4. Potential Measurement Bias: The pattern suggests that the measurement system (the inspectors) may be introducing variation, rather than the manufacturing process itself. This is a serious concern because it means we cannot trust the data to accurately reflect the true quality of the glass.

Recommendation: As a quality control engineer, I would:

• Investigate why defect counts vary by shift
• Provide additional training to ensure consistent inspection standards
• Consider implementing automated defection detection to reduce human variability
• Monitor for inspector fatigue effects

Verification: The pattern analysis is consistent with the data showing systematic variation tied to 8-hour cycles rather than random variation. ✓

Given Information:

• \(\log_{x}(2) = a\)
• \(\log_{x}(5) = b\)
• \(\log_{x}(7) = c\)

Key Logarithm Properties to Use:

1. Product Rule: \(\log_{x}(MN) = \log_{x}(M) + \log_{x}(N)\)
2. Power Rule: \(\log_{x}(M^n) = n\log_{x}(M)\)
3. Quotient Rule: \(\log_{x}(M/N) = \log_{x}(M) - \log_{x}(N)\)

Part a Solution:

Step 1: Factor 10 into numbers we know.

\[ 10 = 2 \times 5 \]

Step 2: Apply the product rule.

\[ \log_{x}(10) = \log_{x}(2 \times 5) \] \[ \log_{x}(10) = \log_{x}(2) + \log_{x}(5) \]

Step 3: Substitute the given values.

\[ \log_{x}(10) = a + b \]

Answer: \(\log_{x}(10) = a + b\)

Verification: Since \(10 = 2 \times 5\), the product rule correctly gives us the sum of the individual logarithms. ✓

Part b Solution:

Step 1: Express 49 as a power.

\[ 49 = 7^2 \]

Step 2: Apply the power rule.

\[ \log_{x}(49) = \log_{x}(7^2) \] \[ \log_{x}(49) = 2\log_{x}(7) \]

Step 3: Substitute the given value.

\[ \log_{x}(49) = 2c \]

Answer: \(\log_{x}(49) = 2c\)

Verification: Since \(49 = 7^2\), the power rule correctly gives us twice the logarithm of 7. ✓

Part c Solution:

Step 1: Factor 50 into numbers we know.

\[ 50 = 2 \times 25 = 2 \times 5^2 \]

Step 2: Apply the product rule.

\[ \log_{x}(50) = \log_{x}(2 \times 5^2) \] \[ \log_{x}(50) = \log_{x}(2) + \log_{x}(5^2) \]

Step 3: Apply the power rule to \(\log_{x}(5^2)\).

\[ \log_{x}(50) = \log_{x}(2) + 2\log_{x}(5) \]

Step 4: Substitute the given values.

\[ \log_{x}(50) = a + 2b \]

Answer: \(\log_{x}(50) = a + 2b\)

Verification: Since \(50 = 2 \times 5^2\), we correctly applied both the product rule and power rule to get \(a + 2b\). ✓

Part d Solution:

Step 1: Factor 56 into numbers we know.

\[ 56 = 8 \times 7 = 2^3 \times 7 \]

Step 2: Apply the product rule.

\[ \log_{x}(56) = \log_{x}(2^3 \times 7) \] \[ \log_{x}(56) = \log_{x}(2^3) + \log_{x}(7) \]

Step 3: Apply the power rule to \(\log_{x}(2^3)\).

\[ \log_{x}(56) = 3\log_{x}(2) + \log_{x}(7) \]

Step 4: Substitute the given values.

\[ \log_{x}(56) = 3a + c \]

Answer: \(\log_{x}(56) = 3a + c\)

Verification: Since \(56 = 2^3 \times 7 = 8 \times 7\), we correctly applied both the product rule and power rule to get \(3a + c\). ✓

Transformation Rules Reference:

For a function \(f(x)\) transformed to \(af(x - h) + k\):

• \(h\): Horizontal shift (positive \(h\) = right, negative \(h\) = left)
• \(k\): Vertical shift (positive \(k\) = up, negative \(k\) = down)
• \(a\): Vertical stretch/compression and reflection
  • \(|a| > 1\): vertical stretch
  • \(0 < |a| < 1\): vertical compression
  • \(a < 0\): reflection over the \(x\)-axis

Important: Inside the function (with \(x\)), the sign is opposite to the direction of the shift.

Part a Solution:

Step 1: Identify the parent function and transformed function.

• Parent function: \(f(x) = \frac{1}{x}\)
• Transformed function: \(g(x) = \frac{1}{x + 2} - 3\)

Step 2: Rewrite in transformation form.

\[ g(x) = \frac{1}{x - (-2)} + (-3) \]

This is in the form \(f(x - h) + k\) where \(h = -2\) and \(k = -3\).

Step 3: Identify each transformation.

1. Horizontal shift: \(x + 2\) means \(x - (-2)\), so \(h = -2\)
   • This shifts the graph 2 units to the LEFT
2. Vertical shift: \(-3\) means \(k = -3\)
   • This shifts the graph 3 units DOWN

Step 4: Describe the complete transformation.

Answer: The graph of \(f(x) = \frac{1}{x}\) is transformed by:

1. Shifting 2 units to the left (horizontal shift)
2. Shifting 3 units down (vertical shift)

This moves the vertical asymptote from \(x = 0\) to \(x = -2\), and the horizontal asymptote from \(y = 0\) to \(y = -3\).

Verification: The center of the hyperbola moves from \((0, 0)\) to \((-2, -3)\). ✓

Part b Solution:

Step 1: Identify the parent function and transformed function.

• Parent function: \(y = x^2\)
• Transformed function: \(y = -(x - 1)^2 + 20\)

Step 2: Identify each transformation.

This is in the form \(y = a(x - h)^2 + k\) where \(a = -1\), \(h = 1\), and \(k = 20\).

1. Horizontal shift: \((x - 1)\) means \(h = 1\)
   • This shifts the graph 1 unit to the RIGHT
2. Vertical shift: \(+20\) means \(k = 20\)
   • This shifts the graph 20 units UP
3. Reflection: The negative sign (\(a = -1\)) means
   • The graph is reflected over the \(x\)-axis (opens downward instead of upward)

Step 3: Describe the complete transformation.

Answer: The graph of \(y = x^2\) is transformed by:

1. Shifting 1 unit to the right (horizontal shift)
2. Shifting 20 units up (vertical shift)
3. Reflecting over the \(x\)-axis (the parabola opens downward)

The vertex moves from \((0, 0)\) to \((1, 20)\), and the parabola opens downward instead of upward.

Verification: The vertex of the original parabola at \((0, 0)\) moves to \((1, 20)\), and the parabola now opens downward. ✓

Part c Solution:

Step 1: Identify the parent function and transformed function.

• Parent function: \(y = |x|\)
• Transformed function: \(y = -3|x + 71|\)

Step 2: Rewrite in transformation form.

\[ y = -3|x - (-71)| \]

This is in the form \(y = a|x - h| + k\) where \(a = -3\), \(h = -71\), and \(k = 0\).

Step 3: Identify each transformation.

1. Horizontal shift: \(x + 71\) means \(x - (-71)\), so \(h = -71\)
   • This shifts the graph 71 units to the LEFT
2. Vertical stretch: \(|-3| = 3\) means the graph is stretched vertically by a factor of 3
   • The graph is stretched vertically by a factor of 3 (steeper)
3. Reflection: The negative sign (\(a = -3 < 0\)) means
   • The graph is reflected over the \(x\)-axis (opens downward instead of upward)
4. Vertical shift: \(k = 0\) (no vertical shift)

Step 4: Describe the complete transformation.

Answer: The graph of \(y = |x|\) is transformed by:

1. Shifting 71 units to the left (horizontal shift)
2. Stretching vertically by a factor of 3 (makes the V-shape steeper)
3. Reflecting over the \(x\)-axis (the V opens downward instead of upward)

The vertex moves from \((0, 0)\) to \((-71, 0)\), the graph is steeper (slope of arms changes from \(\pm 1\) to \(\pm 3\)), and it opens downward.

Verification: The vertex moves from \((0, 0)\) to \((-71, 0)\). For \(x = -70\), we get \(y = -3|{-70} + 71| = -3|1| = -3\), confirming the reflection and stretch. ✓

Step 1: Express both sides using the same base.

We need to rewrite both \(\frac{1}{8}\) and \(\frac{1}{2}\) as powers of the same base. Let's use base 2.

\[ \frac{1}{8} = \frac{1}{2^3} = 2^{-3} \] \[ \frac{1}{2} = 2^{-1} \]

Step 2: Substitute these expressions into the original equation.

\[ \left(2^{-3}\right)^{(2x - 3)} = \left(2^{-1}\right)^{(x + 2)} \]

Step 3: Apply the power rule: \((a^m)^n = a^{mn}\).

\[ 2^{-3(2x - 3)} = 2^{-1(x + 2)} \]

Step 4: Simplify the exponents.

Left side:

\[ -3(2x - 3) = -6x + 9 \]

Right side:

\[ -1(x + 2) = -x - 2 \]

So our equation becomes:

\[ 2^{-6x + 9} = 2^{-x - 2} \]

Step 5: Since the bases are equal, set the exponents equal.

When \(a^m = a^n\) (and \(a \neq 0, 1, -1\)), then \(m = n\).

\[ -6x + 9 = -x - 2 \]

Step 6: Solve for \(x\).

Add \(6x\) to both sides:

\[ 9 = 5x - 2 \]

Add \(2\) to both sides:

\[ 11 = 5x \]

Divide by \(5\):

\[ x = \frac{11}{5} \]

Answer: \(x = \frac{11}{5}\) or \(x = 2.2\)

Verification:

Let's substitute \(x = \frac{11}{5}\) back into the original equation to verify.

Left side:

\[ \left(\frac{1}{8}\right)^{(2x - 3)} = \left(\frac{1}{8}\right)^{(2 \cdot \frac{11}{5} - 3)} \] \[ = \left(\frac{1}{8}\right)^{(\frac{22}{5} - 3)} = \left(\frac{1}{8}\right)^{(\frac{22}{5} - \frac{15}{5})} = \left(\frac{1}{8}\right)^{\frac{7}{5}} \] \[ = \left(2^{-3}\right)^{\frac{7}{5}} = 2^{-\frac{21}{5}} \]

Right side:

\[ \left(\frac{1}{2}\right)^{(x + 2)} = \left(\frac{1}{2}\right)^{(\frac{11}{5} + 2)} \] \[ = \left(\frac{1}{2}\right)^{(\frac{11}{5} + \frac{10}{5})} = \left(\frac{1}{2}\right)^{\frac{21}{5}} \] \[ = \left(2^{-1}\right)^{\frac{21}{5}} = 2^{-\frac{21}{5}} \]

Comparison:

\[ 2^{-\frac{21}{5}} = 2^{-\frac{21}{5}} \quad ✓ \]

Both sides equal \(2^{-\frac{21}{5}}\), confirming our solution is correct!