1
Question
In a triangle ABC, b cosC + c cos B is always less than b + c.
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Solution
The correct option is A True
We know that in a triangle sum of two sides are always more than the third side.
We also know with the help of projection formula b cosC + c cos B = a
So, b + c > a
Or b + c > b cosC + c cos B
Therefore the given statement is true.
We can also verify the given statement in this way -
b cos C≤b;−1≤cos(θ)≤1
Similarly c cos B≤c
Also, b cos C becomes b when C = 0
And c cos B becomes c when C = 0
Which can never be the case in a triangle, therefore
b cos C + c cos B < b + c
We know that in a triangle sum of two sides are always more than the third side.
We also know with the help of projection formula b cosC + c cos B = a
So, b + c > a
Or b + c > b cosC + c cos B
Therefore the given statement is true.
We can also verify the given statement in this way -
b cos C≤b;−1≤cos(θ)≤1
Similarly c cos B≤c
Also, b cos C becomes b when C = 0
And c cos B becomes c when C = 0
Which can never be the case in a triangle, therefore
b cos C + c cos B < b + c
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