Question:
If the lengths of the sides of a triangle a, b, c \( (a>b>c) \) are in arithmetic progression and the greatest angle is twice the smallest, then a: b: \( c= \)
If the lengths of the sides of a triangle a, b, c \( (a>b>c) \) are in arithmetic progression and the greatest angle is twice the smallest, then a: b: \( c= \)
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For multi-variable triangle problems with integer options, checking the options directly using the Law of Cosines is significantly faster than expanding high-degree multi-variable polynomials!
Updated On: Jun 8, 2026
- 6:5:4
- 5:6:7
- 5:4:3
- 13:9:5
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The Correct Option is A
Solution and Explanation
Concept:
Since the side lengths \( a, b, c \) are in an Arithmetic Progression with \( a > b > c \), the middle side satisfies:
\[
2b = a + c \quad \cdots (1)
\]
The largest angle is \( A \) (opposite side \( a \)) and the smallest angle is \( C \) (opposite side \( c \)). We are given that:
\[
A = 2C
\]
Step 1: Applying the Law of Sines to the angle relation.
From the Sine Rule: \( \frac{a}{sin~A} = \frac{c}{sin~C} \). Substitute \( A = 2C \): \[ \frac{a}{sin~2C} = \frac{c}{sin~C} \implies \frac{a}{2sin~C~cos~C} = \frac{c}{sin~C} \implies a = 2c~cos~C \implies cos~C = \frac{a}{2c} \quad \cdots (2) \]
Step 2: Using the Law of Cosines for angle \( C \).
The Law of Cosines states that \( cos~C = \frac{a^2 + b^2 - c^2}{2ab} \). Equating this with equation (2): \[ \frac{a^2 + b^2 - c^2}{2ab} = \frac{a}{2c} \implies c(a^2 + b^2 - c^2) = a^2b \implies ca^2 + cb^2 - c^3 = a^2b \] \[ b^2c - a^2b + ca^2 - c^3 = 0 \] Substitute the A.P. condition \( a = 2b - c \) or check the standard discrete integer options to find which one fits both conditions.
Step 3: Testing standard option integer ratios.
Let us verify Option (A) 6:5:4 where \( a=6, b=5, c=4 \):
• Check A.P. property: \( a + c = 6 + 4 = 10 \). Since \( 2b = 2(5) = 10 \), the A.P. condition is perfectly satisfied.
• Check the cosine angle values: \[ cos~C = \frac{6^2 + 5^2 - 4^2}{2 \times 6 \times 5} = \frac{36 + 25 - 16}{60} = \frac{45}{60} = \frac{3}{4} \] \[ cos~A = \frac{5^2 + 4^2 - 6^2}{2 \times 5 \times 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8} \]
• Test double-angle condition \( cos~A = 2cos^2C - 1 \): \[ 2\left(\frac{3}{4}\right)^2 - 1 = 2\left(\frac{9}{16}\right) - 1 = \frac{9}{8} - 1 = \frac{1}{8} \]
Since \( \frac{1}{8} = \frac{1}{8} \), the condition \( A = 2C \) holds perfectly for the ratio 6:5:4.
Step 1: Applying the Law of Sines to the angle relation.
From the Sine Rule: \( \frac{a}{sin~A} = \frac{c}{sin~C} \). Substitute \( A = 2C \): \[ \frac{a}{sin~2C} = \frac{c}{sin~C} \implies \frac{a}{2sin~C~cos~C} = \frac{c}{sin~C} \implies a = 2c~cos~C \implies cos~C = \frac{a}{2c} \quad \cdots (2) \]
Step 2: Using the Law of Cosines for angle \( C \).
The Law of Cosines states that \( cos~C = \frac{a^2 + b^2 - c^2}{2ab} \). Equating this with equation (2): \[ \frac{a^2 + b^2 - c^2}{2ab} = \frac{a}{2c} \implies c(a^2 + b^2 - c^2) = a^2b \implies ca^2 + cb^2 - c^3 = a^2b \] \[ b^2c - a^2b + ca^2 - c^3 = 0 \] Substitute the A.P. condition \( a = 2b - c \) or check the standard discrete integer options to find which one fits both conditions.
Step 3: Testing standard option integer ratios.
Let us verify Option (A) 6:5:4 where \( a=6, b=5, c=4 \):
• Check A.P. property: \( a + c = 6 + 4 = 10 \). Since \( 2b = 2(5) = 10 \), the A.P. condition is perfectly satisfied.
• Check the cosine angle values: \[ cos~C = \frac{6^2 + 5^2 - 4^2}{2 \times 6 \times 5} = \frac{36 + 25 - 16}{60} = \frac{45}{60} = \frac{3}{4} \] \[ cos~A = \frac{5^2 + 4^2 - 6^2}{2 \times 5 \times 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8} \]
• Test double-angle condition \( cos~A = 2cos^2C - 1 \): \[ 2\left(\frac{3}{4}\right)^2 - 1 = 2\left(\frac{9}{16}\right) - 1 = \frac{9}{8} - 1 = \frac{1}{8} \]
Since \( \frac{1}{8} = \frac{1}{8} \), the condition \( A = 2C \) holds perfectly for the ratio 6:5:4.
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