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Tems11 [23]
3 years ago
11

For Question B Work out the angle BXC GIVE A REASON FOR EACH ANGLE YOU WORK OUT.

Mathematics
1 answer:
babymother [125]3 years ago
5 0
M∠BXC = 70°.

We start out with m∠XBC.  It is 55°, because it is a corresponding angle with ∠AXY.

Since ∠XBC and ∠XCB are the base angles of an isosceles triangle, they are congruent.  This means that m∠XCB = 55°.

To find the measure of ∠BXC, we find the sum of the two base angles in the isosceles triangle and subtract it from 180:

180-(55+55) = 180-110 = 70°
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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)

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4 years ago
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2) All sides of a rhombus are equal.

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