Triangles ABC and DEF are similar find the length of segment DF

Mathematics

1 answer:

asambeis [7]2 years ago

8 0

Answer:

2.68

Step-by-step explanation:

In similar triangles, the corresponding sides are in same ratio

$\frac{DF}{AC} =\frac{DE}{AB}\\\\\frac{DF}{4} = \frac{1.34}{2}\\\\ DF = \frac{1.34}{2}*4\\\\DF = 1.34 * 2 \\\\DF = 2.68$

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2 years ago

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nadya68 [22]

Answer:

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Step-by-step explanation:

$\sf \longrightarrow \frac{ \sec x \sin( - x) + \tan( - x) }{1 + \sec( - x) } \\ \\ \sf \longrightarrow \frac{ \sec x( - \sin x) - \tan x}{1 + \sec x} \\ \\ \sf \longrightarrow \frac{ - \frac{1}{ \cos x } \times \sin x - \tan x }{1 + \sec x} \\ \\ \sf \longrightarrow \frac{ - \tan x - \tan x}{1 + \sec x} \\ \\ \sf \longrightarrow \frac{ - 2 \tan x}{1 + \sec x} \\ \\ \sf \longrightarrow \frac{ \frac{ - 2 \sin x}{ \cos x} }{1 + \frac{1}{ \cos x} } \\ \\ \sf \longrightarrow \frac{ \frac{ - 2 \sin x}{ \cos x} }{ \frac{ \cos x + 1}{ \cos x} } \\ \\ \boxed{ \sf{\longrightarrow \frac{ - 2 \sin x}{ \cos x + 1} }}$