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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.
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
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
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