# Find the Cosine and sine of the angle

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Aufgabe:

A circle of radius 45mm has a chord of length 60mm. Find the sine and the cosine of the angle at the centre of the circle subtended by this chord

Problem/Ansatz:

i don't get it why the ans is 0,994, 0,111. Can anyone explain with a formula pls. I really want to understand, but it makes no sense

Verwende den Kosinussatz. ## 2 Antworten

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Draw a triangle with sides a=45mm, b=45mm and c=60mm.

Draw a line from the midpoint of c to the opposite corner of the triangel (i.e. the height of the triangle).

The angle $$\gamma$$ at that corner satisfies

$$\sin\frac{\gamma}{2} = \frac{60/2}{45}$$

because the height splits the triangle into two congruent right triangles.

vor von
0 Daumen

Hi,

first: make yourself a sketch, and add the symmetry axis (red) Can anyone explain with a formula pls

Don't ask for a formula, look at the sketch and try to remember how sine and cosine are defined.

The angle at $$M$$ (yellow) should be $$\alpha$$. The triangle $$\triangle MBD$$ has a right angle at point $$D$$, so is:$$\sin\left(\frac{\alpha}{2}\right) = \frac{30}{45} = \frac 23$$Have a look at the half-angle formulae and You will find$$\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos(\alpha)}{2}} \\ \implies \cos(\alpha) = 1-2\sin^2\left(\frac{\alpha}{2}\right) = \frac 19 = 0.\overline{1}$$and the value for $$\sin(\alpha)$$ is (ask Pythagoras $$\implies \sin^2+\cos^2=1$$)$$\sin(\alpha) = \sqrt{1-\cos^2(\alpha)} = \frac 19\sqrt{80} = \frac 49\sqrt{5} \approx 0.9938$$

Thank you for ur time, this was so helpful!

Sry, for another question. Today i tried to calculate it on myself. Where does the sqrt(80),sqrt(5) come from?

Where does the sqrt(80),sqrt(5) come from?

\cos(\alpha) = \frac 19 \\ \implies \cos^2(\alpha) = \frac{1}{9^2}= \frac1{81} \\ \begin{aligned}\implies \sqrt{1 - \cos^2(\alpha)} &= \sqrt{1 - \frac1{81}}\\ &= \sqrt{\frac{81}{81} - \frac1{81}}\\&= \sqrt{\frac{81-1}{81}} \\&= \sqrt{\frac{80}{9^2}} \\&= \frac 19\sqrt{80}\\&= \frac19 \sqrt{16 \cdot 5} \\&= \frac19\sqrt{4^2 \cdot 5} \\&= \frac49\sqrt 5\end{aligned}Hint: Practice algebra and fractions

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