Question

deriving a tangent ratio for special triangles
which statements are true regarding triangle lmn? choose three correct answers.
lm = x\sqrt{2}
tan(45°)=\frac{\sqrt{2}}{2}

Explanation:

Step1: Recall Pythagorean theorem for right - isosceles triangle

In right - isosceles triangle LMN with legs LN = x and MN = x. By Pythagorean theorem \(LM^{2}=LN^{2}+MN^{2}\). Substituting LN = x and MN = x, we get \(LM^{2}=x^{2}+x^{2}=2x^{2}\), so \(LM = x\sqrt{2}\).

Step2: Recall tangent ratio formula

The tangent of an angle in a right - triangle is defined as \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\angle L = 45^{\circ}\), \(\tan(45^{\circ})=\frac{MN}{LN}\). Since \(LN = MN=x\), \(\tan(45^{\circ})=\frac{x}{x}=1
eq\frac{\sqrt{2}}{2}\).

Answer:

LM = x√2 is a correct statement. (We need the full list of options to choose three correct answers, but based on the work so far we know this one is correct. If we assume the other non - shown options are evaluated in a similar way using right - triangle trigonometry and Pythagorean theorem principles).