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Bumek [7]
3 years ago
7

How to do question 22?

Mathematics
1 answer:
Aleks04 [339]3 years ago
6 0

Answer:

A = 20sinθ(6 + 5 cosθ)  cm²

Step-by-step explanation:

Drop perpendiculars DE and CF to AB.

Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.

The formula for the area of the trapezium is

A = ½(a + b)h

DE = 10sinθ

AE = 10cosθ

BF = 10cosθ

EF = CD = 12 cm

AB = AE + EF  + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ

A = ½(a + b)h

   = ½(12 +12 + 20 cosθ) × 10 sinθ

   =(24 + 20 cosθ) × 5 sinθ

   = 4(6 + 5cosθ) × 5sinθ

   = 20sinθ(6 + 5 cosθ)  cm²

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Answer:

The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

Step-by-step explanation:

<u><em>Step(i):</em></u>-

Given  

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

                  cos( 3x - \frac{\pi }{3} )  = cos (\frac{\pi }{6} )

                      3x - \frac{\pi }{3}  =  \frac{\pi }{6}

                      3x - \frac{\pi }{3  } + \frac{\pi }{3}   =  \frac{\pi }{6} + \frac{\pi }{3}

                      3x = \frac{2\pi +\pi }{6} = \frac{3\pi }{6} = \frac{\pi }{2}

                     x = \frac{\pi }{6}

<u><em>Step(ii)</em></u>:-

The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

<u><em>verification </em></u>:-

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

put  x = \frac{\pi }{6}

    cos( 3(\frac{\pi }{6})  - \frac{\pi }{3} )  = \frac{\sqrt{3} }{2}

    cos (\frac{\pi }{6} ) = \frac{\sqrt{3} }{2} \\\\\frac{\sqrt{3} }{2} =  \frac{\sqrt{3} }{2}

Both are equal

∴The solution of the given trigonometric equation

                   x = \frac{\pi }{6}

                     

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