Question

A right triangle has perimeter of length $7$ and hypotenuse of length $3$ if $θ$ is the largest non right angle in the triangle then value of $\cos (θ)$ equals to :

$a) \: \frac{ \sqrt{6} - \sqrt{2} }{4}$
$b) \: \frac{4 + \sqrt{2} }{6}$
$c) \: \frac{4 - \sqrt{2} }{3}$
$d) \: \frac{4 - \sqrt{2} }{6}$
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Answer 1

Answer: cos∅ = (4 + √2) / 6

Step-by-step explanation:

The hypotenuse = 3 units

Perimeter = 7 units

Sum of lengths of the other two sides = perimeter - hypotenuse = 7 - 3 = 4 units

Let one side be x, the other side = 4 - x

Using Pythagoras theorem, we get

=> (x)² + (4-x)² = 3²

=> x² + 16 + x² - 8x = 9

=> 2x² - 8x + 7 = 0

Using the quadratic equation formula, we can find the values of x which satisfy the equation

Quadratic formula = [(-b) ± √(b² - 4ac)] / 2a = x

Where a, b, c are values is in the equation ax² + bx + c = 0

Using this, we get a = 2, b = -8, c = 7

Substituting the values in the quadratic formula, we get

=> x = [8 ± √(8² - 4*2*7)] / 2*2 = [8 ± √(64 - 56)] / 4 = [8 ± 2√2] / 4

=> x = [4 ± √2] / 2

Thefore the sides are (4 + √2)/2 and (4 - √2)/2

Cos∅ = (4+√2)/2] / 3 = 4+√2 / 6

Thefore cos∅ = (4 + √2) / 6

Please brainlist my answer, if helpful!