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How do you solve this trigonometry problem?
DEF is a triangle in which DE=DF=17cm and EF=16cm. Find the lengths of the heights DM and EN where DM and EN are perpendicular to EF and DF respectively.

Sagot :

hint: draw the triangle first
DM=sq.root of (17cm^2 - 8cm^2)                        pythagorean theorem
note: 8cm comes from half of 16 which is EF
        ^2 means to the power of 2

EN:
let DN=x
    NF=17-x
solving for EN you got 2 equations:
EN=sq.root of(17cm^2 - x^2)
&
EN=sq.root of(16cm^2 - {17cm-x}^2)

equate 2 EN and you get:
sq.root of(17^2-x^2)=sq.root of(16^2-{17-x}^2)
17^2-x^2={16^2}-{17-x}^2                                  both sides are squared
289 - x^2 = 256 - 289 +34x - x^2
34x = 322
x = 161/17

therefore EN = 17^2 - x^2
                   = 289 - (161/17)^2
                   =240/17

Ihartangelidn
First, to analyze this..you should illustrate the problem..

solve first for DM
since DE and DF is equal then DM must me cutting EF into two equal parts, EM=EF=8
since it would be a right triangle then we can use the Pythagorean theorem in whinc hyp^2=a^2+b^2
threfore substituting, 17^2=8^2+(EM)^2
EM=15

i"m sorry i have no idea on finding EN..that was all i know