Question

In \( \triangle ABC \), if \( b + c : c + a : a + b = 7:8:9 \), then the smallest angle (in radians) of that triangle is:

• A. \( \cos^{-1} \left( \frac{4}{5} \right) \)
• B. \( \frac{\pi}{3} \)
• C. \( \cos^{-1} \left( \frac{3}{5} \right) \)
• D. \( \frac{\pi}{4} \)

Hint:

For any triangle, the smallest angle corresponds to the smallest side. The cosine rule: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] helps in computing angles efficiently.

Answer 1

The Correct Option is A

Step 1: Understanding the Given Ratio The given condition: \[ b + c : c + a : a + b = 7:8:9 \] Using the sum property in a triangle: \[ (b + c) + (c + a) + (a + b) = 2(a + b + c). \] Setting \( 7x, 8x, 9x \) as the respective values: \[ 7x + 8x + 9x = 2(a + b + c). \]
Step 2: Finding Side Ratios Solving for \( a + b + c \): \[ a + b + c = \frac{24x}{2} = 12x. \] Now, express the sides as: \[ a = \frac{(8x + 9x - 7x)}{2} = \frac{10x}{2} = 5x, \] \[ b = \frac{(7x + 9x - 8x)}{2} = \frac{8x}{2} = 4x, \] \[ c = \frac{(7x + 8x - 9x)}{2} = \frac{6x}{2} = 3x. \] Thus, the sides are in the ratio: \[ a : b : c = 5 : 4 : 3. \]
Step 3: Applying Cosine Rule to Find the Smallest Angle Since \( c = 3x \) is the smallest side, we use the cosine rule: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab}. \] Substituting \( a = 5x \), \( b = 4x \), and \( c = 3x \): \[ \cos C = \frac{(5x)^2 + (4x)^2 - (3x)^2}{2(5x)(4x)}. \] \[ \cos C = \frac{25x^2 + 16x^2 - 9x^2}{2(20x^2)} = \frac{32x^2}{40x^2} = \frac{4}{5}. \]
Step 4: Finding the Smallest Angle \[ C = \cos^{-1} \left( \frac{4}{5} \right). \] Thus, the smallest angle is: \[ \boxed{\cos^{-1} \left( \frac{4}{5} \right).} \]

Answer 2

In triangle \( \triangle ABC \), we are given that the ratios of the sums of sides are \( b+c : c+a : a+b = 7:8:9 \). Let's express these sums in terms of a common parameter \( k \):

\( b+c = 7k \), \( c+a = 8k \), and \( a+b = 9k \).

Now, let's add these equations:

\( (b+c) + (c+a) + (a+b) = 7k + 8k + 9k = 24k \)

Which simplifies to:

\( 2(a+b+c) = 24k \Rightarrow a+b+c = 12k \)

Let's express each side in terms of \( a, b, \) and \( c \):

\( b = 12k - (8k) = 4k \)

\( c = 12k - (9k) = 3k \)

\( a = 12k - (7k) = 5k \)

The sides of the triangle are \( a = 5k \), \( b = 4k \), and \( c = 3k \).

Now we need to determine the smallest angle. By the law of cosines:

\( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \)

Plug in the values:

\( (3k)^2 = (5k)^2 + (4k)^2 - 2(5k)(4k)\cos(C) \)

\( 9k^2 = 25k^2 + 16k^2 - 40k^2\cos(C) \)

\( 9 = 25 + 16 - 40\cos(C) \)

\( 9 = 41 - 40\cos(C) \)

\( 40\cos(C) = 32 \)

\( \cos(C) = \frac{4}{5} \)

This implies the smallest angle \( C \) is:

\( C = \cos^{-1} \left( \frac{4}{5} \right) \)

Thus, the smallest angle in radians is \( \cos^{-1} \left( \frac{4}{5} \right) \).