Question

Given: KLMN is a trapezoid, KL = MN

LN = 89 ,
LH
- altitude
LM = 3, KN = 13.
Find: m∠LKN

Answer 1

LKN = 90 degrees

Let's solve this with a different approach:

LM = 3, KN = 13 --> Given

HN = (KN + LM) / 2 = (13+3)/2 = 16/2 = 8

LN =
`√(89)` --> Given

`LH^(2)` =
`√(89) ^(2) - HN^(2)` --> Pythagorean Theorem

We get: LH = 5

Similarly,

KH = (KN - LM) / 2 = (13-3)/2 = 10/2 = 5

We now see that triangle LHK is isosceles - both sides are equal

KHL = 90 degrees --> Def. of altitude

That leaves angle HLK = angle LKN = 90/2 = 45 degrees --> Base angle theorem.

LKN = 90 degrees

Answer 2

m∠LNK = 86.8°

Explanation:

∵ LH ⊥ KN at H

∵ LK = MN

∵ LM = 3

∴ HN = (13 - 3) ÷ 2 + 3 = 8 ⇒ isosceles trapezium

∵ LN = 89

∴ LH = √(89² - 8²) = 88.64

∵ KH = (13 - 3) ÷ 2 = 5 ⇒ isosceles trapezium

∵ tan ∠LKN = LH/KH = 88.64/5

∴ m∠LKN = 86.77° ≅ 86.8°