Question

If in a triangle, \[a{\cos ^2}\dfrac{C}{2} + c{\cos ^2}\dfrac{A}{2} = \dfrac{{3b}}{2}\], then its sides will be in
A. A.P.
B. G.P.
C. H.P.
D. A.G.

Answer 1

Hint: The distributive property states that if you divide something among several things, each thing gets an equal share of the divisor. In this case, since \[a{\cos ^2}\dfrac{C}{2} + c{\cos ^2}\dfrac{A}{2} = \dfrac{{3b}}{2}\].
Simply compute \[s\left( {a - b} \right)\] using the Pythagorean Theorem and then subtract \[\;csc\left( a \right) + csi\] to find that side (b). This means that one side (side A) of the triangle is in algebraic formula P and the other two sides are in algebraic formula Q.

Complete step by step solution: The given equation is \[a{\cos ^2}\dfrac{C}{2} + c{\cos ^2}\dfrac{A}{2} = \dfrac{{3b}}{2}\].
This equation can also be written as,
\[a\dfrac{{s(s - c)}}{{ab}} + c\dfrac{{s(s - a)}}{{bc}} = \dfrac{{3b}}{2}\]
After solving, the equation becomes,
\[2s(s - c + s - a)\]
Which then becomes,
\[3{b^2}\]
\[2s(b) = 3{b^2}\]
By taking b in common, the equation becomes
\[2s = 3b\]
The appropriate structure for the equation is,
\[a + b + c = 3b\]
When the common term b, goes on the same side, the equation becomes,
\[a + c = 2b\]
Hence, the triangle a, b, c is in A.

So, Option ‘A’ is correct

Note: \[a{\cos ^2}\dfrac{C}{2} + c{\cos ^2}\dfrac{A}{2} = \dfrac{{3b}}{2}\]then it is best to reach out to a mathematician. While there may be an equation that could hide those values within its confines, solving that equation would likely fall outside of the scope of this question.
Algebra and geometry will most certainly play a role in such a solution - something which cannot be easily explained or conveyed in simple prose.
The same outcome is obtained by multiplying the sum of two or more addends by a number as it is by multiplying each addend by the number separately and combining the resulting products.