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ohaa [14]
1 year ago
8

Triangle JKL is similar to triangle MNO

Mathematics
1 answer:
astra-53 [7]1 year ago
5 0

Answer:

Perimeter is 20 and x = 10

Step-by-step explanation:

MNO is 2/3 of JKL. So x = 15 x 2/3.  x = 10. 10 + 4 + 6 = 20

Hope this helps!

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I need help proving this ASAP
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Answer:

See explanation

Step-by-step explanation:

We want to show that:

\tan(x +  \frac{3\pi}{2} )  =  -   \cot \: x

One way is to use the basic double angle formula:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x)  \cos( \frac{3\pi}{2} )  +   \cos(x)  \sin( \frac{3\pi}{2}) }{\cos(x)  \cos( \frac{3\pi}{2} )   -    \sin(x)  \sin( \frac{3\pi}{2}) }

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ \sin(x) ( 0)  +   \cos(x) (  - 1) }{\cos(x) (0)   -    \sin(x) (  - 1) }

We simplify further to get:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{ 0  -   \cos(x) }{0 +    \sin(x) }

We simplify again to get;

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  \frac{- \cos(x) }{ \sin(x) }

This finally gives:

\frac{ \sin(x +  \frac{3\pi}{2} ) }{\cos(x +  \frac{3\pi}{2} )}  =  -  \cot(x)

6 0
3 years ago
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3 0
1 year ago
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6 0
3 years ago
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