Question

deriving a tangent ratio for special triangles
which statements are true regarding triangle lmn?
check all that apply.
☐ nm = x
☐ nm = ( xsqrt{2} )
☐ lm = ( xsqrt{2} )
☐ ( \tan(45^circ) = \frac{sqrt{2}}{2} )
☐ ( \tan(45^circ) = 1 )

Explanation:

Step1: Analyze Triangle Type

Triangle \( LMN \) is a right - isosceles triangle (since \( \angle L=\angle M = 45^{\circ} \), so \( \angle N = 90^{\circ} \), and \( LN=NM \) as it is an isosceles right triangle). Given \( LN=x \), so \( NM = x \) (because in an isosceles right triangle, the two legs are equal).

Step2: Calculate Hypotenuse \( LM \)

Using the Pythagorean theorem \( c^{2}=a^{2}+b^{2} \) for right - triangle \( LMN \) where \( a = LN=x \), \( b = NM=x \) and \( c = LM \). Then \( LM^{2}=x^{2}+x^{2}=2x^{2} \), so \( LM=\sqrt{2x^{2}}=x\sqrt{2} \).

Step3: Analyze Tangent of \( 45^{\circ} \)

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

Answer:

• \( \text{NM}=x \) (True)
• \( \text{LM}=x\sqrt{2} \) (True)
• \( \tan(45^{\circ}) = 1 \) (True)