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Question

If $tan invisible function application A plus tan invisible function application B plus tan invisible function application C equals 3 √ 3$ then the $capital delta to the power of k$ is

isosceles
right angled
equilateral
scalene

hint Hint:

In this question we have to use trigonometric formula $tan open parentheses A close parentheses plus tan open parentheses B close parentheses plus tan open parentheses C close parentheses equals tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses$ . Then we will Calculate A.M and G.M and check whether they are equal or not. If they are equal then this is only possible when all the terms are equal.

The correct answer is: equilateral

In this question we have to find the type of triangle
We are given $tan open parentheses A close parentheses plus tan open parentheses B close parentheses plus tan open parentheses C close parentheses equals 3 square root of 3$
We know that in a triangle
$A plus B plus C space equals pi$
$A plus B equals pi minus C$
$tan open parentheses A plus B close parentheses equals tan open parentheses pi minus C close parentheses$
=> $tan open parentheses A plus B close parentheses equals negative tan open parentheses C close parentheses$
=> $tan open parentheses A plus B close parentheses equals fraction numerator tan open parentheses A close parentheses plus tan open parentheses B close parentheses over denominator 1 minus tan open parentheses A close parentheses tan open parentheses B close parentheses end fraction$
=> $fraction numerator tan open parentheses A close parentheses plus tan open parentheses B close parentheses over denominator 1 minus tan open parentheses A close parentheses tan open parentheses B close parentheses end fraction equals negative tan open parentheses C close parentheses$
=> $tan open parentheses A close parentheses plus tan open parentheses B close parentheses equals negative tan open parentheses C close parentheses left parenthesis 1 minus tan open parentheses A close parentheses tan open parentheses B close parentheses right parenthesis$
=> $tan open parentheses A close parentheses plus tan open parentheses B close parentheses plus tan open parentheses C close parentheses equals tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses$
=> We know that $A. space M space greater or equal than space G. M$
=> $fraction numerator tan open parentheses A close parentheses plus tan open parentheses B close parentheses plus tan open parentheses C close parentheses over denominator 3 end fraction greater or equal than left parenthesis tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses right parenthesis to the power of 1 third end exponent$
=> $fraction numerator 3 square root of 3 over denominator 3 end fraction greater or equal than left parenthesis tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses right parenthesis to the power of 1 third end exponent$
=> $square root of 3 greater or equal than left parenthesis tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses right parenthesis to the power of 1 third end exponent$
=> $tan open parentheses A close parentheses tan open parentheses B close parentheses tan open parentheses C close parentheses less or equal than 3 square root of 3$
Here we are getting $A. M space equals G. M$ ,
It is possible when $tan open parentheses A close parentheses equals tan open parentheses B close parentheses equals tan open parentheses C close parentheses$
=> $A equals B equals C$
All the angles are equal.
Hence, the given triangle is equilateral triangle

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