What is the equivalent trig ratio to cos 18 ?

Mathematics

1 answer:

Anika [276]3 years ago

3 0

Answer:

$\sin{72^\circ}$

Step-by-step explanation:

In a right triangle, the two acute angles are complementary--they add up to 90 degrees.

Cosine is defined as the adjacent side divided by the hypotenuse. In the attached picture, $\cos{18^\circ}=\frac{a}{h}$ .

The other acute angle is 90 - 18 = 72 degrees. Looking at just that angle, side <em>a</em> is the opposite side. So the ratio a/h is opposite over hypotenuse -- the sine ratio.

Therefor, $\cos{18^\circ} = \sin{72^\circ}$

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F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$