Question

In triangle ABC, if \( a = 4, b = 3, c = 2 \), then \( 2(a - b \cos C)(a - c \sec B) = \):

• A. 0
• B. 1
• C. 2
• D. 3 \bigskip

Hint:

For solving trigonometric expressions in triangles, the Law of Cosines is essential for determining unknown angles or sides. Always simplify expressions carefully and validate your steps using known trigonometric identities.

Answer 1

The Correct Option is D

We are given a triangle ABC with the following values: \[ a = 4, \quad b = 3, \quad c = 2. \] Our objective is to compute \( 2(a - b \cos C)(a - c \sec B) \). 

Step 1: Apply the Law of Cosines to determine \( \cos C \) and \( \cos B \). The Law of Cosines states: \[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] Substituting the given values: \[ \cos C = \frac{4^2 + 3^2 - 2^2}{2 \times 4 \times 3} = \frac{16 + 9 - 4}{24} = \frac{21}{24} = \frac{7}{8}. \] 

 Step 2: Similarly, using the Law of Cosines to find \( \cos B \): \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] Substituting the values: \[ \cos B = \frac{4^2 + 2^2 - 3^2}{2 \times 4 \times 2} = \frac{16 + 4 - 9}{16} = \frac{11}{16}. \] 

Step 3: Given that \( \cos C = \frac{7}{8} \) and \( \cos B = \frac{11}{16} \), we proceed with evaluating \( 2(a - b \cos C)(a - c \sec B) \). Since \( \sec B \) is the reciprocal of \( \cos B \): \[ \sec B = \frac{1}{\cos B} = \frac{1}{\frac{11}{16}} = \frac{16}{11}. \] 

Step 4: Now compute \( 2(a - b \cos C)(a - c \sec B) \): \[ a - b \cos C = 4 - 3 \times \frac{7}{8} = 4 - \frac{21}{8} = \frac{32}{8} - \frac{21}{8} = \frac{11}{8}, \] \[ a - c \sec B = 4 - 2 \times \frac{16}{11} = 4 - \frac{32}{11} = \frac{44}{11} - \frac{32}{11} = \frac{12}{11}. \] Now, multiply these results: \[ (a - b \cos C)(a - c \sec B) = \frac{11}{8} \times \frac{12}{11} = \frac{12}{8} = \frac{3}{2}. \] Multiplying by 2: \[ 2(a - b \cos C)(a - c \sec B) = 2 \times \frac{3}{2} = 3. \] Thus, the final result is: \[ \boxed{3}. \]