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2 years ago

18 cm

Mathematics

1 answer:

attashe74 [19]2 years ago

5 0

Answer

720π cm²

Step-by-step explanation:

Surface area of cylinder

A=2πr(s+r)

for r=12cm, s=18cm

A= 2π×12(12+18)

=24π×30

=720π

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3 years ago

Find the exact value of 6cos(105°)

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Answer:

$-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}$

Step-by-step explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of $6\cos(105^{\circ}})$ is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

<u>Solution 1 (Cosine Addition Identity):</u>

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:

$\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta$

Notice that $45+60=105$ and therefore we can easily solve this problem if we know values of $\cos(45^{\circ})$ , $\cos(60^{\circ})$ , $\sin (45^{\circ})$ , and $\sin(60^{\circ})$ , which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:

$\cos(45^{\circ})=\sin(45^{\circ})=\frac{\sqrt{2}}{2}$

$\cos(60^{\circ})=\frac{1}{2}$
$\sin(60^{\circ})=\frac{\sqrt{3}}{2}$

Therefore, we have:

$\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}$

Since we want the value of $6\cos 105^{\circ}$ , simply multiply this by 6 to get your final answer:

$6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}$

<u>Solution 2 (Combination of trig. identities):</u>

Although less plausible, you may have the following memorized:

$\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$

If so, we can use the following trig. identity:

$\cos(\theta)=\sin(90^{\circ}-\theta)$ (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,

$\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})$

Recall another trig. identity:

$\sin(-\theta)=-\sin (\theta)$ and therefore:

$\sin (-15^{\circ})=-\sin (15^{\circ})$

Multiply by 6 to get:

$6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}$ (alternative final answer).

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