Raul's family is buying a house on a an odd-shaped lot. The lengths of three of the sides are given by the expressions shown. Th

e remaining two sides lengths are equal to each other. The perimeter of the lot is given by the expression 13y-5. Which expression gives the length of the remaining side?

Mathematics

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Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Leona [12618]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

To determine: KN and KL

1. Analyzing triangle NLM, we see it is a right triangle due to NL ⊥ LM. In this context,

LM = 45

m∠M = 50°

Consequently,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

It is also true that

(angles LNM and M are complementary). $m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$

2. Now considering triangle NKL, it also forms a right triangle as KN ⊥ NM. Within this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

Thus,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$