Question

Given A, B, C represents angles of a  ⃤ AB and cosA + 2 cosB + cosC = 2 and AB = 3 and BC = 7 then cosA – cosC is?

• A. \(-\frac{10}{7}\)
• B. \(\frac{10}{7}\)
• C. \(\frac{5}{7}\)
• D. \(-\frac{5}{7}\)

Answer 1

The Correct Option is A

The given equation is:

\( \cos A + \cos C = 2(1 - \cos B) \)

Rewriting the left-hand side using the sum-to-product identities:

\( 2 \cos \frac{A + C}{2} \cos \frac{A - C}{2} = 4 \sin^2 \frac{B}{2} \)

From the identity \( \cos \frac{A+C}{2} = \sin \frac{B}{2} \), we have:

\( \cos \frac{A-C}{2} = 2 \sin \frac{B}{2} \)

Substituting back:

\( 2 \cos \frac{B}{2} \cos \frac{A-C}{2} = 4 \sin \frac{B}{2} \cos \frac{B}{2} \)

Simplify further:

\( 2 \sin \frac{A+C}{2} \cos \frac{A-C}{2} = 4 \sin \frac{B}{2} \cos \frac{B}{2} \)

This shows:

\( \sin A + \sin C = 2 \sin B \)

Given \( a + c = 2b \), substitute \( a = 3, c = 7 \), and \( b = 5 \).

Now compute:

\( \cos A - \cos C = \frac{b^2 + c^2 - a^2}{2bc} - \frac{a^2 + b^2 - c^2}{2ab} \)

Substitute the values:

\( \cos A - \cos C = \frac{25 + 49 - 9}{70} - \frac{9 + 25 - 49}{30} \)

Simplify:

\( \cos A - \cos C = \frac{65}{70} - \frac{-15}{30} \)

Convert to a common denominator:

\( \cos A - \cos C = \frac{65}{70} + \frac{35}{70} = \frac{100}{70} = \frac{10}{7}\)

Simplify further:

\( \cos A - \cos C = \frac{10}{7} \)

Answer 2

The correct answer is option (A) \(-\frac{10}{7}\)