Answer
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Hint: Use Law of cosines and also know the basic properties of the triangle like a triangle cannot be both acute and obtuse. First of all, a triangle cannot be obtuse and acute at the same time. In this particular case, the triangle is acute.
The side we wish to find the length; I do believe you meant hypotenuse, not hypothesis. However, only right triangles have hypotenuses, which are the sides opposite of the right angle. Also, it is somewhat incorrect to call the other sides "base" and "vertical side", due to this triangle not being right angled.
Complete step by step solution:
Anyway, here's our triangle:
Now, the Law of Cosines states that, if a and b are two sides of a triangle and the angle between them is θ, then the third side, denoted c here, is
\[
{c^2} = {a^2} + {b^2} - 2ab\cos \theta \\
{x^2} = {2^2} + {4^2} - 2 \times 2 \times 4\cos {75^ \circ } \\
\]
In the above equation we substitute the values as:
$
c = x \\
a = 2 \\
b = 4 \\
\theta = {75^ \circ } \\
$
We can calculate $\cos 75$ using sum identities:
Here we use this formula to get the value of cos 75 $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Writing 75 as 30+45
Hence, we have:
$
\cos ({30^ \circ } + {45^ \circ }) = \cos {30^ \circ } \times \cos {45^ \circ } - \sin {30^ \circ } \times \sin {45^ \circ } \\
= \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 2 }}{2} - \dfrac{1}{2} \times \dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} \\
\therefore {x^2} = 4 + 16 - 16 \times (\dfrac{{\sqrt 6 - \sqrt 2 }}{4}) \\
\therefore {x^2} = 20 - 4\sqrt 6 + 4\sqrt 2 \\
x = \sqrt {20 - 4\sqrt 6 + 4\sqrt 2 } = 2\sqrt {5 - \sqrt 6 + \sqrt 2 } \\
$
Note: There are some things to clear up. The way you phrased your question, it was a bit confusing. First of all, a triangle cannot be obtuse and acute at the same time. In this particular case, the triangle is acute. And the formulas like the identities and law of cosine and sine are important so make sure to remember them well.
The side we wish to find the length; I do believe you meant hypotenuse, not hypothesis. However, only right triangles have hypotenuses, which are the sides opposite of the right angle. Also, it is somewhat incorrect to call the other sides "base" and "vertical side", due to this triangle not being right angled.
Complete step by step solution:
Anyway, here's our triangle:
Now, the Law of Cosines states that, if a and b are two sides of a triangle and the angle between them is θ, then the third side, denoted c here, is
\[
{c^2} = {a^2} + {b^2} - 2ab\cos \theta \\
{x^2} = {2^2} + {4^2} - 2 \times 2 \times 4\cos {75^ \circ } \\
\]
In the above equation we substitute the values as:
$
c = x \\
a = 2 \\
b = 4 \\
\theta = {75^ \circ } \\
$
We can calculate $\cos 75$ using sum identities:
Here we use this formula to get the value of cos 75 $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Writing 75 as 30+45
Hence, we have:
$
\cos ({30^ \circ } + {45^ \circ }) = \cos {30^ \circ } \times \cos {45^ \circ } - \sin {30^ \circ } \times \sin {45^ \circ } \\
= \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 2 }}{2} - \dfrac{1}{2} \times \dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 6 - \sqrt 2 }}{4} \\
\therefore {x^2} = 4 + 16 - 16 \times (\dfrac{{\sqrt 6 - \sqrt 2 }}{4}) \\
\therefore {x^2} = 20 - 4\sqrt 6 + 4\sqrt 2 \\
x = \sqrt {20 - 4\sqrt 6 + 4\sqrt 2 } = 2\sqrt {5 - \sqrt 6 + \sqrt 2 } \\
$
Note: There are some things to clear up. The way you phrased your question, it was a bit confusing. First of all, a triangle cannot be obtuse and acute at the same time. In this particular case, the triangle is acute. And the formulas like the identities and law of cosine and sine are important so make sure to remember them well.
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