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hjlf
3 years ago
12

Given WXYZ is a rhombus.

Mathematics
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

Diagonal of rhombus bisect the angles of it.

Step-by-step explanation:

Given: WXYZ is a rhombus.

To prove: WY bisects Angle ZWX and Angle XYZ.  ZX bisects Angle WZY and Angle YXW.

In rhombus WXYZ, WY is a diagonal.

In triangle WXY and triangle WZY,

WX=WZ        (sides of rhombus are equal)

XY=ZY        (sides of rhombus are equal)

WY=WY       (Common side)

By SSS postulate,

\triangle WXY\cong \triangle WZY

\angle WYX\cong \angle WYZ       (CPCTC)

\angle XWY\cong \angle ZWY       (CPCTC)

Hence prove, that WY bisects Angle ZWX and Angle XYZ.

Similarly,

In rhombus WXYZ, ZX is a diagonal.

In triangle WXZ and triangle YXZ,

WX=YX        (sides of rhombus are equal)

XZ=XZ         (Common side)

WZ=YZ        (sides of rhombus are equal)

By SSS postulate,

\triangle WXZ\cong \triangle YXZ

\angle WZX\cong \angle YZX       (CPCTC)

\angle WXZ\cong \angle YXZ       (CPCTC)

Hence prove, that ZX bisects Angle WZY and Angle YXW.

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Step-by-step explanation:

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First, let's look at the identities:

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Add.

sin(A - B) = \frac{4}{5}

This is your answer.

Hope this helps!

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