In a triangle ABC, if acos^2C/2+ccos^2A/2=3b/2, then the sides of the triangle are in

A. AP. Step 1: With cos^2C/2=1+cosC/2: a+c/2+acosC+ccosA/2=3b/2. Step 2: Projection rule: acosC+ccosA=b. So a+c+b=3b, i.e. a+c=2b. Step 3: Hence a, b, c are in AP. Check: Sides a=3, b=5, c=7 ( a+c=2b): the left side works out to 7.5=3× 5/2. Trap: (B) GP fails: for sides 4, 6, 9 ( b^2=ac) the left side is not 3b/2=9.

Question 9 of 15 · Properties of Triangles

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NIMCET 2021

In a triangle ABC, if $a cos^{2} \frac{C}{2} + c cos^{2} \frac{A}{2} = \frac{3 b}{2}$ , then the sides of the triangle are in

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — AP

Detailed Solution

With $cos^{2} \frac{C}{2} = \frac{1 + cos C}{2}$ : $\frac{a + c}{2} + \frac{a cos C + ccos A}{2} = \frac{3 b}{2}$ .

Projection rule: $a cos C + ccos A = b$ . So $a + c + b = 3 b$ , i.e. $a + c = 2 b$ .

Hence $a$ , $b$ , $c$ are in AP.

✓ Check

Sides

a = 3

b = 5

c = 7

(

a + c = 2 b

): the left side works out to

7.5 = \frac{3 \times 5}{2}

⚠ Common trap

(B) GP fails: for sides 4, 6, 9 (

b^{2} = a c

) the left side is not

\frac{3 b}{2} = 9

Quick tip

$2 cos^{2} \frac{C}{2} = 1 + cos C$ and projection rule $a cos C + c cos A = b \Rightarrow a + c = 2 b$

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Properties of Triangles — Preparation

This question tests your understanding of Properties of Triangles, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Properties of triangles connect the sides a, b, c and the opposite angles A, B, C of any triangle, not just right-angled ones. This topic covers the sine rule (a/sin A = b/sin B = c/sin C = 2R), the cosine rule (a^2 = b^2 + c^2 - 2bc cos A), the projection formulas, half-angle formulas, and formulas for the area of a triangle, including (1/2)ab sin C and Hero's formula. It also covers the circumradius R, the inradius r and their relationships with the area and semi-perimeter, such as r = area/s. Exam questions test applying these rules to find an unknown side, angle or area, and identifying the type of triangle from a given condition. Choosing between the sine rule and the cosine rule is usually the key decision. Practise the properties of triangles questions below, each with a step-by-step solution.

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