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Tcecarenko [31]
4 years ago
5

If θ is an angle in standard position whose terminal side passes through (3, 4), evaluate tan(1/2)θ.

Mathematics
1 answer:
aleksandr82 [10.1K]4 years ago
7 0

The tangent half angle formula, one of several, is

\tan \dfrac a 2 = \dfrac{1 - \cos a}{\sin a}

We have θ is opposite 4 in the 3/4/5 right triangle so

\cos \theta = \dfrac{3}{5}

\sin \theta = \dfrac{4}{5}

\tan \dfrac{\theta}{2} = \dfrac{1 - 3/5}{4/5} = \dfrac{5-3}{4}=\dfrac{1}{2}

Answer: 1/2

This is actually pretty deep.  It says half the big acute angle in the 3/4/5 triangle is the small diagonal angle of the 1x2 rectangle.   Similarly, the small acute angle in 3/4/5 triangle is twice the small diagonal angle of the 1x3 rectangle.

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The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

y = ax^2 + bx + c

The given quadratic equation is 24x^2 + 2x = 15

Let's rearrange the equation

24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0

This implies that

  • a = 24
  • b = 2
  • c = -15

The equation or formula of quadratic formula is given as

y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}

We can substitute the values into the equation and solve

y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}

From the calculations above, the solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

Learn more on quadratic equation here;

brainly.com/question/8649555

#SPJ1

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