Answer:
DL = 5.87 un.
LN = 10.98 un.
DN = 14.4 un.
Step-by-step explanation:
Given:
∆DLN,
LF ⊥ DN ,
DF = 4
m∠N = 23º,
m∠D = 47º
Find:
DL, LN, DN
Solution:
In right triangle DFL, DF = 4, m∠D = 47º, so
[tex]LF=DF\tan \angle D\\ \\LF=4\cdot \tan 47^{\circ}\approx 4.29 \ units[/tex]
and
[tex]DL=\dfrac{DF}{\cos \angle D}\\ \\DL=\dfrac{4}{\cos 47^{\circ}}\approx 5.87\ units[/tex]
In right triangle LFN, LF = 4.29 units, m∠N = 23º, so
[tex]FN=LF\cot \angle N\\ \\FN=4.29\cdot \cot 23^{\circ}\approx 10.11\ units[/tex]
and
[tex]LN=\dfrac{LF}{\sin \angle N}\\ \\LN=\dfrac{4.29}{\sin 23^{\circ}}\approx 10.98\ units[/tex]
Hence,
DN = 4.29 + 10.11 = 14.4 units
