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Tems11 [23]
3 years ago
5

1. ∠XVR =

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

Answer/Step-by-step explanation:

1. ∠XVR = 180 - <XVW (angle on a straight line)

∠XVR = 180 - 55°

∠XVR = 125°

2. ∠RVS = <XVW (Vertical angles are congruent)

∠RVS = 55°

3. ∠WVS = ∠XVR (vertical angles are congruent)

∠WVS = 125°

4. ∠RST = <R + <RVS (exterior angle theorem)

<RST = 55 + 55

<RST = 110°

5. ∠RSV = 180 - (<R + <RVS) (sum of triangle)

∠RSV = 180 - (55 + 55)

∠RSV = 70°

6. ∠VSU = <RST (vertical angles are congruent)

<VSU = 70°

7. ∠UST = <RSV (vertical angles)

<UST = 70°

8. ∠TUS = 180 - (<UST + <T) (sum of triangle)

<TUS = 180 - (70 + 71)

<TUS = 39°

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In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th
olya-2409 [2.1K]

Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


8 0
3 years ago
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