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Olenka [21]
3 years ago
9

Which expression is equivalent to (tan(θ/2))(sin θ)

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
8 0

Answer:

B) -1+cos\theta

Step-by-step explanation:

(tan(\frac{\theta}{2}))(-sin\theta)

(\frac{1-cos\theta}{sin\theta})(-sin\theta)

(\frac{-sin\theta+sin\theta cos\theta}{sin\theta})

-1+cos\theta

Therefore, option B is correct

<u>Helpful Tips:</u>

Half-Angle Formula: tan(\frac{\theta}{2})= \frac{1-cos\theta}{sin\theta}=\frac{sin\theta}{1+cos\theta}=\pm\sqrt{\frac{1+cos\theta}{2}}

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Help please! 10 pts!<br> if mCED=62 what is x?
irinina [24]

Answer:

The value of x^{0} is 59^{0}.

Step-by-step explanation:

The figure provided to is a rectangle, named ABDC.

All the angles, m∠CAB = m∠ABD= m∠BDC = m∠DCA = 90^{0}.

The lines AD and BC are diagonals of the rectangle ABDC.

According to the diagonal property of rectangles, they bisect each other.

Then,

  • m∠CED = m∠AEB and m∠BED = m∠AEC
  • AE = BE, CE = ED
  • The opposite angle at the points where the diagonals meet are congruent, i.e. m∠DAB = m∠ADC and  m∠DCB = m∠ABC.

Now, consider the triangle CED.

Since the triangle CED has two equal sides, i.e. CE = ED, it is an isosceles triangle. And hence the angles m∠DCE = m∠EDC = a^{0} (say).

Compute the value of m∠DCE and m∠EDC using the sum of angles property of a triangle i.e. the sum of all three angles of a triangle is 180^{0}.

Solve for a^{0} as follows:

m∠CED + m∠ECD + m∠EDC = 180^{0}

62^{0} + a^{0} + a^{0} = 180^{0}

62^{0}+2a^{0}=180^{0}\\2a^{0}=180^{0}-62^{0}  \\a^{0}=\frac{118^{0} }{2}\\=59^{0}

So, m∠EDC = 59^{0} = m∠ADC.

As the opposite angles at the points where the diagonals meet are congruent, then,

m∠DAB = m∠ADC = 59^{0}.

Thus, the value of x^{0} is 59^{0}.

6 0
3 years ago
Read 2 more answers
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