In a right angled triangle One acute angle is double the other then the hypotenuse is
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but how do u prove this?
Let the smallest acute angle be x . Then,
9 0 ∘ + x + 2 x = 1 8 0 ∘
Solving this, we get x = 3 0 ∘ .
This problem is the exact statement of 3 0 ∘ − 6 0 ∘ − 9 0 ∘ theorem which states that if in a triangle, the angles are 3 0 ∘ , 6 0 ∘ , & 9 0 ∘ , then the hypotenuse is twice the shortest side.
but how do u prove this?
If we solve a simultaneous equation that is related to the angle of the right triangle, then we get x + y = 1 8 0 − 9 0 , x = 2 y .
Then, we get x = 6 0 , y = 3 0 .
This problem is related to trigonometric functions, but we do not have to know that.
If each angle of the triangle is 3 0 , 6 0 , 9 0 , then the length of the hypotenuse is double of its shortest side.
Then, the correct answer is Double the smallest side .
PS: It is not the 'smallest'. In the geometrical works, we don't say 'the smallest' because the side cannot be the smallest. We say 'shortest' instead.
Apply the 30-60-90 special right triangle
"the longer leg is square root of 3 times as long as the shorter leg while the hypotenuse is twice as long as the shorter leg"
but how do u prove this?
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It is the 30-60-90 triangle with ratio 1 : 3 : 2