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Hitman42 [59]
3 years ago
6

Suppose that

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

sin(B) = 12/13

cos(B) = 5/13

tan(B) = 12/5

csc(B) = 13/12

sec(B) = 13/5

cot(B) = 5/12

Step-by-step explanation:

If ABC is a right triangle, assuming that ∠C = 90°, then the segment AB =13 is the hypotenuse and the other two sides are:

BC = 5\\AC = \sqrt{13^2 - 5^2}\\AC = 12

The six  trigonometric functions of angle B are:

sin(B) =\frac{AC}{AB}= \frac{12}{13}\\cos(B) = \frac{BC}{AB} =\frac{5}{13}\\tan(B) = \frac{AC}{BC} =\frac{12}{5}\\\csc(B) =\frac{1}{sin(B)}= \frac{13}{12}\\sec(B) = \frac{1}{cos(B)} =\frac{13}{5}\\cot(B) = \frac{1}{tan(B)} =\frac{5}{12}\\

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<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

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x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16

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Now plug in values in eqn 1, we get,

x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}

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I’ll name brainliest✨
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