Respuesta :
Answer:
A.) NM= x
C.) LM = x√2
E.) tan (45°) = 1
Step-by-step explanation:
If the legs are both x, then the hypotenuse is equal to [tex]x\sqrt{2[/tex]
Therefore, LM= [tex]x\sqrt{2[/tex] is correct and MN= x
Disclaimer: The sum is done according to the picture attached as the question given is wrong.
The true statements regarding the given isosceles right triangle ΔLMN are:
- NM = x
- LM = x√2
- tan 45° = 1.
What are isosceles right triangles?
An isosceles triangle is a triangle where two sides and their corresponding angles are equal.
A right triangle is a triangle with one angle = 90°.
An isosceles right triangle is a right-angled triangle with two legs including the right angle are equal. Their corresponding angles are equal and each of them = 45°. So, the three angles of an isosceles right triangle are 45°, 45°, and 90°, always.
How do we solve the given question?
In the figure, we can see that we have a ΔLMN, with ∠L = 45°, ∠M = 45°, and ∠N = 90°. Also, we can see that LN = x.
The given angles of ΔLMN determine that it is an isosceles right triangle with a right angle at N.
Since, the two legs involving the right angle, that is N, are equal, we can say that, NM = LN = x.
The hypotenuse of the ΔLMN, that is the side opposite to ∠N, that is LM, can be found using the Pythagoras theorem, by which in a right-angled triangle,
Hypotenuse² = Base² + Perpendicular².
∴ LM² = LN² + NM² = x² + x² = 2x².
or, LM = √(2x²) = x√2.
The tangent of an angle ∅, that is, tan ∅ is computed using the formula,
tan ∅ = Perpendicular/Base.
To calculate tan 45°, that is, tangent to ∠L, we take Perpendicular = NM and Base = LN.
∴ tan 45° = NM/LN = x/x = 1.
Now, we check all the given options:
- NM = x. TRUE (computed)
- NM = x√2. FALSE (∵ NM = x)
- LM = x√2. TRUE (computed)
- tan 45° = √2/2. FALSE (∵ tan 45° = 1)
- tan 45° = 1. TRUE (computed)
∴ The true statements regarding the given isosceles right triangle ΔLMN are:
- NM = x
- LM = x√2
- tan 45° = 1.
Learn more about isosceles right triangle at
https://brainly.com/question/691225
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