2 years ago

The two trapezoids in the graph below are similar. Which sequence of transformations proves that the trapezoids are similar?

Mathematics

2 answers:

antiseptic1488 [7]2 years ago

8 0

Answer:

I think the answer is B

Hoochie [10]2 years ago

3 0

The answer is B.

Hopes this helps!! :')

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Sin5A/SinA-cos5A/cosA=4cos2A

irina1246 [14]

Answer:

See Explanation

Step-by-step explanation:

$\frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ LHS = \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} \\ \\ = \frac{ \sin5A \:\cos A - \cos5A \: \sin A}{\sin A \:\cos A } \\ \\ = \frac{ \sin(5A -A )}{\sin A \:\cos A} \\ \\ = \frac{ \sin 4A}{\sin A \:\cos A} \\ \\ = \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ =4\cos 2A \\ \\ = RHS \\ \\ thus \\ \\ \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ hence \: proved$