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Salsk061 [2.6K]
3 years ago
10

Find three consecutive positive integers such that the sum of their squares is 2354. What is the largest integer?

Mathematics
1 answer:
Ilya [14]3 years ago
4 0

Answer:

The largest integer is 29.

Step-by-step explanation:

Let the consecutive positive integers are x, x+1 and x+2.

The square of sum of squares of three consecutive positive integers is 2354 such that,

x^2+(x+1)^2+(x+2)^2=2354\\\\x^2+x^2+2x+1+x^2+4+4x=2354\\\\3x^2+6x+5=2354\\\\3x^2+6x- 2349=0

It is a quadratic equation whose solution is given by :

x = 27 and x = -29

First positive integer = 27

Second positive integer = 27+1 = 28

Third positive integer = 27+2 = 29

Hence, the largest integer is 29.

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Half Angle Formula

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Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

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