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Question

In triangle $A B C comma fraction numerator a over denominator b end fraction equals fraction numerator 2 over denominator 3 end fraction blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction.$ Then the number of triangles satisfying these conditions is

0
1
2
3

The correct answer is: 2

We have $blank fraction numerator a over denominator 2 end fraction equals fraction numerator b over denominator 3 end fraction equals k blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction$
$rightwards double arrow cos to the power of 2 end exponent invisible function application A equals fraction numerator 5 over denominator 8 end fraction$
$rightwards double arrow fraction numerator 5 over denominator 8 end fraction equals open parentheses fraction numerator 9 k to the power of 2 end exponent plus c to the power of 2 end exponent minus 4 k to the power of 2 end exponent over denominator 6 k c end fraction close parentheses to the power of 2 end exponent equals open parentheses fraction numerator 5 k to the power of 2 end exponent plus c to the power of 2 end exponent over denominator 6 k c end fraction close parentheses to the power of 2 end exponent$
$rightwards double arrow 45 k to the power of 2 end exponent c to the power of 2 end exponent equals 50 k to the power of 4 end exponent plus 20 k to the power of 2 end exponent c to the power of 2 end exponent plus 2 c to the power of 4 end exponent$
$rightwards double arrow 2 c to the power of 4 end exponent minus 25 k to the power of 2 end exponent c to the power of 2 end exponent plus 50 k to the power of 4 end exponent equals 0$
$rightwards double arrow c to the power of 2 end exponent equals fraction numerator 25 k to the power of 2 end exponent plus-or-minus square root of 625 k to the power of 4 end exponent minus 400 blank k to the power of 4 end exponent end root over denominator 4 end fraction$
$rightwards double arrow fraction numerator 25 k to the power of 2 end exponent plus-or-minus 15 k to the power of 2 end exponent over denominator 4 end fraction equals 10 k to the power of 2 end exponent comma fraction numerator 5 over denominator 2 end fraction blank k to the power of 2 end exponent$
There are two possible valid values of $c to the power of 2 end exponent.$ Hence, there exist two triangles satisfying the given conditions

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Maths-General

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