Answer :
Let's address the given problems step-by-step:
### Part b:
Problem: In triangle [tex]\(ABC\)[/tex], where [tex]\(A = 100^\circ\)[/tex], and [tex]\(4A = 24B\)[/tex], find angle [tex]\(C\)[/tex] in radians.
1. Determine angle [tex]\( B \)[/tex]:
- We know [tex]\( A = 100^\circ \)[/tex].
- According to the given relationship: [tex]\( 4A = 24B \)[/tex].
- Substituting [tex]\(A\)[/tex] into the equation: [tex]\( 4 \times 100^\circ = 24B \)[/tex].
- This simplifies to [tex]\( 400^\circ = 24B \)[/tex].
- Solving for [tex]\(B\)[/tex]: [tex]\( B = \frac{400^\circ}{24} \)[/tex].
- Therefore, angle [tex]\( B = 16.6667^\circ \)[/tex] (approximately [tex]\(16.67^\circ\)[/tex]).
2. Find angle [tex]\( C \)[/tex]:
- In any triangle, the sum of the interior angles is always [tex]\(180^\circ\)[/tex].
- Therefore, [tex]\( A + B + C = 180^\circ \)[/tex].
- Substituting [tex]\( A \)[/tex] and [tex]\( B \)[/tex]: [tex]\( 100^\circ + 16.6667^\circ + C = 180^\circ \)[/tex].
- Solving for [tex]\(C\)[/tex]: [tex]\( C = 180^\circ - 100^\circ - 16.6667^\circ \)[/tex].
- Therefore, angle [tex]\( C = 63.3333^\circ \)[/tex] (approximately [tex]\(63.33^\circ\)[/tex]).
3. Convert angle [tex]\( C \)[/tex] to radians:
- To convert degrees to radians, use the formula: [tex]\( \text{radians} = \text{degrees} \times \frac{\pi}{180} \)[/tex].
- Substituting [tex]\(C\)[/tex]: [tex]\( 63.3333^\circ \times \frac{\pi}{180} \approx 1.1054 \)[/tex] radians.
So, the result for part b is:
- Angle [tex]\( B \approx 16.67^\circ \)[/tex]
- Angle [tex]\( C \approx 63.33^\circ \)[/tex]
- Angle [tex]\( C \)[/tex] in radians [tex]\(\approx 1.1054\)[/tex].
### Part c:
Problem: In triangle [tex]\( XYZ \)[/tex], where [tex]\(2X = 3Y = 6Z\)[/tex], find the angles in degrees.
1. Express the angles in terms of a single variable [tex]\( k \)[/tex]:
- Let [tex]\(2X = k\)[/tex].
- Therefore, [tex]\(X = \frac{k}{2}\)[/tex].
- Similarly, [tex]\(3Y = k \implies Y = \frac{k}{3}\)[/tex].
- And [tex]\(6Z = k \implies Z = \frac{k}{6}\)[/tex].
2. Sum of the angles in a triangle:
- The sum of the angles in any triangle is [tex]\(180^\circ\)[/tex], so [tex]\(X + Y + Z = 180^\circ\)[/tex].
- Substituting the expressions for [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex]:
[tex]\[ \frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ. \][/tex]
3. Solve for [tex]\( k \)[/tex]:
- Find the common denominator for the fractions, which is 6:
[tex]\[ \frac{3k}{6} + \frac{2k}{6} + \frac{k}{6} = 180^\circ. \][/tex]
- Combine the fractions:
[tex]\[ \frac{3k + 2k + k}{6} = 180^\circ. \][/tex]
- Simplify:
[tex]\[ \frac{6k}{6} = 180^\circ. \][/tex]
- Hence, [tex]\(k = 180^\circ\)[/tex].
4. Calculating angles:
- [tex]\(X = \frac{k}{2} = \frac{180^\circ}{2} = 90^\circ\)[/tex].
- [tex]\(Y = \frac{k}{3} = \frac{180^\circ}{3} = 60^\circ\)[/tex].
- [tex]\(Z = \frac{k}{6} = \frac{180^\circ}{6} = 30^\circ\)[/tex].
So, the angles in triangle [tex]\(XYZ\)[/tex] are:
- Angle [tex]\( X = 90^\circ \)[/tex].
- Angle [tex]\( Y = 60^\circ \)[/tex].
- Angle [tex]\( Z = 30^\circ \)[/tex].
### Part b:
Problem: In triangle [tex]\(ABC\)[/tex], where [tex]\(A = 100^\circ\)[/tex], and [tex]\(4A = 24B\)[/tex], find angle [tex]\(C\)[/tex] in radians.
1. Determine angle [tex]\( B \)[/tex]:
- We know [tex]\( A = 100^\circ \)[/tex].
- According to the given relationship: [tex]\( 4A = 24B \)[/tex].
- Substituting [tex]\(A\)[/tex] into the equation: [tex]\( 4 \times 100^\circ = 24B \)[/tex].
- This simplifies to [tex]\( 400^\circ = 24B \)[/tex].
- Solving for [tex]\(B\)[/tex]: [tex]\( B = \frac{400^\circ}{24} \)[/tex].
- Therefore, angle [tex]\( B = 16.6667^\circ \)[/tex] (approximately [tex]\(16.67^\circ\)[/tex]).
2. Find angle [tex]\( C \)[/tex]:
- In any triangle, the sum of the interior angles is always [tex]\(180^\circ\)[/tex].
- Therefore, [tex]\( A + B + C = 180^\circ \)[/tex].
- Substituting [tex]\( A \)[/tex] and [tex]\( B \)[/tex]: [tex]\( 100^\circ + 16.6667^\circ + C = 180^\circ \)[/tex].
- Solving for [tex]\(C\)[/tex]: [tex]\( C = 180^\circ - 100^\circ - 16.6667^\circ \)[/tex].
- Therefore, angle [tex]\( C = 63.3333^\circ \)[/tex] (approximately [tex]\(63.33^\circ\)[/tex]).
3. Convert angle [tex]\( C \)[/tex] to radians:
- To convert degrees to radians, use the formula: [tex]\( \text{radians} = \text{degrees} \times \frac{\pi}{180} \)[/tex].
- Substituting [tex]\(C\)[/tex]: [tex]\( 63.3333^\circ \times \frac{\pi}{180} \approx 1.1054 \)[/tex] radians.
So, the result for part b is:
- Angle [tex]\( B \approx 16.67^\circ \)[/tex]
- Angle [tex]\( C \approx 63.33^\circ \)[/tex]
- Angle [tex]\( C \)[/tex] in radians [tex]\(\approx 1.1054\)[/tex].
### Part c:
Problem: In triangle [tex]\( XYZ \)[/tex], where [tex]\(2X = 3Y = 6Z\)[/tex], find the angles in degrees.
1. Express the angles in terms of a single variable [tex]\( k \)[/tex]:
- Let [tex]\(2X = k\)[/tex].
- Therefore, [tex]\(X = \frac{k}{2}\)[/tex].
- Similarly, [tex]\(3Y = k \implies Y = \frac{k}{3}\)[/tex].
- And [tex]\(6Z = k \implies Z = \frac{k}{6}\)[/tex].
2. Sum of the angles in a triangle:
- The sum of the angles in any triangle is [tex]\(180^\circ\)[/tex], so [tex]\(X + Y + Z = 180^\circ\)[/tex].
- Substituting the expressions for [tex]\(X\)[/tex], [tex]\(Y\)[/tex], and [tex]\(Z\)[/tex]:
[tex]\[ \frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ. \][/tex]
3. Solve for [tex]\( k \)[/tex]:
- Find the common denominator for the fractions, which is 6:
[tex]\[ \frac{3k}{6} + \frac{2k}{6} + \frac{k}{6} = 180^\circ. \][/tex]
- Combine the fractions:
[tex]\[ \frac{3k + 2k + k}{6} = 180^\circ. \][/tex]
- Simplify:
[tex]\[ \frac{6k}{6} = 180^\circ. \][/tex]
- Hence, [tex]\(k = 180^\circ\)[/tex].
4. Calculating angles:
- [tex]\(X = \frac{k}{2} = \frac{180^\circ}{2} = 90^\circ\)[/tex].
- [tex]\(Y = \frac{k}{3} = \frac{180^\circ}{3} = 60^\circ\)[/tex].
- [tex]\(Z = \frac{k}{6} = \frac{180^\circ}{6} = 30^\circ\)[/tex].
So, the angles in triangle [tex]\(XYZ\)[/tex] are:
- Angle [tex]\( X = 90^\circ \)[/tex].
- Angle [tex]\( Y = 60^\circ \)[/tex].
- Angle [tex]\( Z = 30^\circ \)[/tex].
