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3

Two circles are externally tangent. Common tangents to the circles form an angle of measure 2x. How do you prove the ratio of the radii of the circles is (1 - sinx)/(1+ sinx)?

Shwetank Mauria
Featured 1 week ago

Explanation:

Let us have the two circles centered at $P$ and $Q$, with radius ${R}_{1}$ and ${R}_{2}$ respectively.

Let the two tangents be $L M N$ and $G H N$, where $L , M , G , H$ are the points at which they touch the circles as shown and $N$ is the point at which they intersect each other.

The line joining $P Q$ will also pass through $N$. Now join points of contact of tangents $L , M , G , H$ with the respective centers of the circles as shown. Let $m \angle L N G = 2 x$, then $m \angle L N P = x$.

Now $\Delta s L N P$ and $M N Q$ are right angles and hence

${R}_{1} = P N \sin x$ .....................(A)

and ${R}_{2} = Q N \sin x$ .....................(B)

Subtracting (B) from (A), we get

${R}_{1} - {R}_{2} = \left(P N - Q N\right) \sin x$

or $\sin x = \frac{{R}_{1} - {R}_{2}}{P Q} = \frac{{R}_{1} - {R}_{2}}{{R}_{1} + {R}_{2}}$

or $\frac{1}{\sin} x = \frac{{R}_{1} + {R}_{2}}{{R}_{1} - {R}_{2}}$

Applying componendo and dividendo

$\frac{1 + \sin x}{1 - \sin x} = \frac{{R}_{1} + {R}_{2} + {R}_{1} - {R}_{2}}{{R}_{1} + {R}_{2} - {R}_{1} + {R}_{2}}$

$\frac{1 + \sin x}{1 - \sin x} = \frac{2 {R}_{1}}{2 {R}_{2}}$

or ${R}_{2} / {R}_{1} = \frac{1 - \sin x}{1 + \sin x}$

3

Assuming that the Earths diameter is 6336 km and that subtends an angle of 56' at the centre of the moon, find the distance of the moon from the earth. Expain with figure clearly?

dk_ch
Featured 1 week ago

Comparing with the available data of radius of the earth I assume that earth's radius color(blue)(R= 6336" " km subtends an angle $\textcolor{red}{\theta = 56 '}$ at the center of the moon as shown in above figure. Let $D$ be the distance between center of the earth and moon.

$D$ being very large we can assume that the length of the earth's radius $R$ lye on the arc of a circle drawn taking $D$ as radius and moon's center as the center of the circle.

So we can write

"Arc R"/"Radius D"= theta ("in radian")

$\theta = 56 ' = {\left(\frac{56}{60}\right)}^{\circ} = \frac{56}{60} \times \frac{\pi}{180}$radian

So we get

$\frac{R}{D} = \frac{56}{60} \times \frac{\pi}{180}$

$\implies D = \frac{60 \times 180 \times R}{56 \pi} = \frac{60 \times 180 \times 6336}{56 \pi}$km

$\implies D \approx 388956$ km

( It is comparable to available figure as shown below)

2

sin(A-B)=1/2 and cos(A+B)=1/2, then find A and B?

Douglas K.
Featured 1 week ago

Here is a plot of $\textcolor{red}{\sin \left(A - B\right) = \frac{1}{2}}$ and color(blue)(cos(A+B)=1/2

Given:

$\sin \left(A - B\right) = \frac{1}{2} \text{ [1]}$
$\cos \left(A + B\right) = \frac{1}{2} \text{ [2]}$

Using the corresponding inverse on each equation:

$A - B = {\sin}^{-} 1 \left(\frac{1}{2}\right) \text{ [1.1]}$
$A + B = {\cos}^{-} 1 \left(\frac{1}{2}\right) \text{ [2.1]}$

Because the inverses have two possible values, each of the two equations separates into two more equations and each repeats at integer multiples of $2 \pi$:

A-B = pi/6+2n_1pi; n_1 in ZZ" [1.2a]"
A-B = (5pi)/6+2n_1pi; n_1 in ZZ" [1.2b]"
A+B = pi/3+2n_2pi; n_2 in ZZ" [2.2a]"
A+B = (5pi)/3+ 2n_2pi; n_2 in ZZ" [2.2b]"

We have 4 combinations ,$\left(a , a\right) , \left(a , b\right) , \left(b , a\right) , \mathmr{and} \left(b , b\right)$.

Combination $\left(a , a\right)$

Add equation [1.2a] to equation [2.2a]:

2A = pi/6+pi/3+2n_1pi+2n_2pi; n_1 in ZZ,n_2 in ZZ

A = pi/12+pi/6+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ

A = pi/4+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(a,a)]

Subtract equation [1.2a] from equation [2.2a]:

2B = pi/3-pi/6- 2n_1pi+2n_2pi; n_1 in ZZ,n_2 in ZZ

B = pi/6-pi/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ

B = pi/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(a,a)]

The editor is complaining about how long my answer is getting so I shall decrease the amount of detail.

Combination $\left(a , b\right)$

A = (11pi)/12+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(a,b)]

B = (3pi)/4- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(a,b)]

Combination $\left(b , a\right)$

A = (7pi)/12+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(b,a)]

B = -pi/4- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(b,a)]

Combination $\left(b , b\right)$

A = (5pi)/4+n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[3_(b,b)]

B = (5pi)/12- n_1pi+n_2pi; n_1 in ZZ,n_2 in ZZ" "[4_(b,b)]

3

How do you solve tan x - sin 3x = 1?

Sean
Featured 1 week ago

$x = - 2.87192 , 0.930628 , \mathmr{and} 3.41126$

Explanation:

.

The most practical way to solve this equation is by using a graphing utility, visually determining the approximate roots, and then employing an iterative process to get as close as possible to the exact roots. Let's take a look at the graph of one period of the function:

$\tan x - \sin 3 x - 1 = 0$

As you can see, we have three roots that are close to:

$x = - 3 , 1 , \mathmr{and} 3.5$

By trying values to the left and right of each, we can get to the actual values with great accuracy.

But if we wanted to actually solve for $x$ using algebra, we could do it as follows:

$\tan x - 1 = \sin 3 x$

$\tan x - 1 = \sin \left(2 x + x\right)$

Using the following identity:

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$, we get:

$\sin \frac{x}{\cos} x - 1 = \sin 2 x \cos x + \cos 2 x \sin x$

Using the identities:

$\sin 2 x = 2 \sin x \cos x$, and

$\cos 2 x = 2 {\cos}^{2} x - 1$, we get:

$\frac{\sin x - \cos x}{\cos} x = 2 \sin x {\cos}^{2} x + \left(2 {\cos}^{2} x - 1\right) \sin x$

$\frac{\sin x - \cos x}{\cos} x = 2 \sin x {\cos}^{2} x + 2 \sin x {\cos}^{2} x - \sin x$

$\frac{\sin x - \cos x}{\cos} x = 4 \sin x {\cos}^{2} x - \sin x$

Multiplying both sides by $\cos x$, we get:

$\sin x - \cos x = 4 \sin x {\cos}^{3} x - \sin x \cos x$

Rearranging terms, we get:

$\cos x = \sin x \left(1 + \cos x - 4 {\cos}^{3} x\right)$

Solving for $\sin x$, we get:

sinx=cosx/(1+cosx-4cos^3x

In order to solve for $x$, we need to convert this equation into one in terms of either $\sin x$ or $\cos x$. We will go with $\cos x$. To do so, we will square both sides of the equation:

${\sin}^{2} x = {\cos}^{2} \frac{x}{1 + \cos x - 4 {\cos}^{3} x} ^ 2$

Using the identity:

${\sin}^{2} x + {\cos}^{2} x = 1$

We can solve for ${\sin}^{2} x$ and plug it in:

${\sin}^{2} x = 1 - {\cos}^{2} x$

$1 - {\cos}^{2} x = {\cos}^{2} \frac{x}{1 + \cos x - 4 {\cos}^{3} x} ^ 2$

Multiplying both sides by ${\left(1 + \cos x - 4 {\cos}^{3} x\right)}^{2}$, we get:

$\left(1 - {\cos}^{2} x\right) {\left(1 + \cos x - 4 {\cos}^{3} x\right)}^{2} = {\cos}^{2} x$

Let's figure out what ${\left(1 + \cos x - 4 {\cos}^{3} x\right)}^{2}$ is equal to:

${\left(1 + \cos x - 4 {\cos}^{3} x\right)}^{2} = \left(1 + \cos x - 4 {\cos}^{3} x\right) \left(1 + \cos x - 4 {\cos}^{3} x\right)$

After foiling and combining like terms, we end up with:

${\left(1 + \cos x - 4 {\cos}^{3} x\right)}^{2} = 16 {\cos}^{6} x - 8 {\cos}^{4} x - 8 {\cos}^{3} x + {\cos}^{2} x + 2 \cos x + 1$

Let's plug this in:

$\left(1 - {\cos}^{2} x\right) \left(16 {\cos}^{6} x - 8 {\cos}^{4} x - 8 {\cos}^{3} x + {\cos}^{2} x + 2 \cos x + 1\right) - {\cos}^{2} x = 0$

After foiling and combining like terms, we get:

$16 {\cos}^{8} x - 24 {\cos}^{6} x - 8 {\cos}^{5} x + 9 {\cos}^{4} x + 10 {\cos}^{3} x + {\cos}^{2} x - 2 \cos x - 1 = 0$

For simplicity, let's let $\cos x = w$. Our equation turns into:

$16 {w}^{8} - 24 {w}^{6} - 8 {w}^{5} + 9 {w}^{4} + 10 {w}^{3} + {w}^{2} - 2 w - 1 = 0$

This is a polynomial of degree $8$. We can either use the Rational Zeros Theorem, get a list of possible roots in the form of $\frac{p}{q}$, and try them to see if we can find a real root. After finding one, we can perform long division to get an equation of degree $7$. Repeating this process we can potentially get all possible roots.

The other method is the Newton-Raphson method which says:

${x}_{n + 1} = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}$

Start out with ${x}_{0} = 1$ and compute ${x}_{n + 1}$ in an iterative manner until $\Delta {x}_{n + 1} < 0.000001$

Continuing this process, you can find the roots. It is a lengthy process and would exceed what Socratic would allow me to input. But you can look it up in calculus books and follow it.

Back to our equation, it turns out it has $3$ real roots and $4$ complex roots. The complex roots are where the graph has relative maximum, relative minimum and inflection points which are points where the graph does not cross the $x$-axis.

The real roots are where it crosses the $x$-axis. From either of the two methods mentioned above, we can get:

$w = \cos x = \left(- 0.96386 , 0.59733 , \mathmr{and} \ldots \ldots \ldots .\right)$

Within one period of the function, we get three real values for $\cos x$. Calculating the $\arccos$ of each, we get:

$x = - 2.87192 , 0.930628 , \mathmr{and} 3.41126$

which are very close to what we observed at the beginning from the graph. You can try each of these values in the original problem equation and see that they work.

6

How do you find min/max of y=(sinx-cosx+sqrt(2))/(sinx+cosx+2) ?

dk_ch
Featured 1 week ago

$y = \frac{\sin x - \cos x + \sqrt{2}}{\sin x + \cos x + 2}$

=>y=(sqrt2(1/sqrt2sinx-1/sqrt2cosx+1))/(sqrt2(1/sqrt2sinx+1/sqrt2cosx+sqrt2)

$\implies y = \frac{\cos \left(\frac{\pi}{4}\right) \sin x - \sin \left(\frac{\pi}{4}\right) \cos x + 1}{\cos \left(\frac{\pi}{4}\right) \sin x + \sin \left(\frac{\pi}{4}\right) \cos x + \sqrt{2}}$

$\implies y = \frac{\sin \left(x - \frac{\pi}{4}\right) + 1}{\sin \left(x + \frac{\pi}{4}\right) + \sqrt{2}}$

We know $- 1 \le \sin \left(x - \frac{\pi}{4}\right) \le 1$
and $- 1 \le \sin \left(x + \frac{\pi}{4}\right) \le 1$

when $\sin \left(x - \frac{\pi}{4}\right) = - 1$

The minimum value of numerator is =0,

Denominator is always $> 0$

So ${y}_{\text{min}} = 0$

When $\sin \left(x - \frac{\pi}{4}\right) = 1 = \sin \left(2 n \pi + \frac{\pi}{2}\right)$

$\implies x = 2 n \pi + \frac{3 \pi}{4}$
Then

$\sin \left(x + \frac{\pi}{4}\right) = \sin \left(2 n \pi + \frac{3 \pi}{4} + \frac{\pi}{4}\right) = \sin \left(\left(2 n + 1\right) \pi\right) = 0$

So $y = \frac{1 + 1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

When

$\sin \left(x + \frac{\pi}{4}\right) = - 1$

The value of denominator $\sqrt{2} - 1$

Then $\sin \left(x + \frac{\pi}{4}\right) = - 1 = \sin \left(\left(2 n + 1\right) \pi + \frac{\pi}{2}\right)$

$\implies x = \left(2 n + 1\right) \pi + \frac{\pi}{2} - \frac{\pi}{4}$

so $\sin \left(x - \frac{\pi}{4}\right) = \sin \left(\left(2 n + 1\right) \pi + \frac{\pi}{2} - \frac{\pi}{4} - \frac{\pi}{4}\right) = \sin \left(\left(2 n + 1\right) \pi\right) = 0$

then numerator $= 1$

And $y = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$

Hence $y = \sqrt{2} + 1$

Graph shows that ${y}_{\text{min}} = 0$ satisfies the graphical solution. But ${y}_{\text{max}}$ obtained from graph is $\approx 2 \sqrt{2}$ cannot be obtained by the method adopted here.

4

How are hyperbolic sine, hyperbolic cosine, and hyperbolic tangent used in real life?

Rhys
Featured 1 week ago

Catenary Curve...

Explanation:

This is an idea i have heard about a while ago...

Lets say we take a piece of slack string, and hold it, so it is still slack, and so it hangs vertically downward, or even a peice of chain, tethered to two posts, and hangs downward, both of these can be directly modelled by the hyperbolic cosine, $\cosh x$

Or in general the function is: $y = a \cosh \left(\frac{x}{a}\right)$

So in this image, it models a peice of string bieng held at $\left(- 16.4 , 29.88\right) \mathmr{and} \left(16.4 , 29.88\right)$

This is the curve: $y = 8.5 \cosh \left(\frac{x}{8.5}\right)$

Where this curve can be parametrically defined as:

$x \left(t\right) = t$
$y \left(t\right) = a \cosh \left(\frac{t}{a}\right)$

Where also arcs and archways can also be modelled by the caternary curve, where this has the properties of having a very strong foundation and infrastucture...

I hope this is a small insight to how hyperbolic functions can be used in the real world!

3

Can somebody solve it for me?

Rhys
Featured 4 days ago

Proof by conventional trigonometric identities...

Explanation:

We can prove the follwing statment by trig identities...

color(red)( cos^4 theta -= 1/8 ( cos4theta + 4cos2theta + 3 ) ,AA theta

$\forall \theta$ - Meaning "For all theta "

We can color(blue) ( cos^2 theta - sin^2 theta -= cos2theta

$\implies 2 {\cos}^{2} \theta - 1 \equiv \cos 2 \theta$

We can use this identity...

$\implies {\cos}^{2} \theta \equiv \frac{1}{2} \left(\cos 2 \theta + 1\right)$

${\cos}^{4} \theta = {\left({\cos}^{2} \theta\right)}^{2}$

$\implies {\cos}^{4} \theta = {\left(\frac{1}{2} \left(\cos 2 \theta + 1\right)\right)}^{2}$

$\implies {\cos}^{4} \theta = \frac{1}{4} \left({\cos}^{2} 2 \theta + 2 \cos 2 \theta + 1\right)$

Again we can use the fact that color(red)(cos^2 theta = 1/2(cos2theta +1)

=> color(blue)(cos^2 2theta = 1/2 (cos4theta +1 )

$\text{Hence this: } {\cos}^{4} \theta = \frac{1}{4} \left(\textcolor{b l u e}{{\cos}^{2} 2 \theta} + 2 \cos 2 \theta + 1\right)$

$\text{Becomes this: } {\cos}^{4} \theta = \frac{1}{4} \left(\textcolor{b l u e}{\frac{1}{2} \left(\cos 4 \theta + 1\right)} + 2 \cos 2 \theta + 1\right)$

$\implies {\cos}^{4} \theta \equiv \frac{1}{4} \left(\frac{1}{2} \cos 4 \theta + 2 \cos 2 \theta + \frac{3}{2}\right)$

$\implies {\cos}^{4} \theta \equiv \frac{1}{4} \cdot \frac{1}{2} \left(\cos 4 \theta + 4 \cos 2 \theta + 3\right)$

color(red)( => cos^4 theta -= 1/8 ( cos4theta + 4cos2theta + 3 )

2

If r_1=r_2+r_3+r prove that the triangle is right angled.?

dk_ch
Featured yesterday

We know

${r}_{1} = \frac{\Delta}{s - a}$

${r}_{2} = \frac{\Delta}{s - b}$

${r}_{3} = \frac{\Delta}{s - c}$

$r = \frac{\Delta}{s}$

Inserting these values in the given relation.

${r}_{1} = {r}_{2} + {r}_{3} + r$

$\implies {r}_{1} - r = {r}_{2} + {r}_{3}$

$\implies \frac{\Delta}{s - a} - \frac{\Delta}{s} = \frac{\Delta}{s - b} + \frac{\Delta}{s - c}$

$\implies \frac{1}{s - a} - \frac{1}{s} = \frac{1}{s - b} + \frac{1}{s - c}$

$\implies \frac{2}{2 s - 2 a} - \frac{2}{2 s} = \frac{2}{2 s - 2 b} + \frac{2}{2 s - 2 c}$

$\implies \frac{1}{b + c - a} - \frac{1}{b + c + a} = \frac{1}{c + a - b} + \frac{1}{a + b - c}$

$\implies \frac{\left(b + c + a\right) - \left(b + c - a\right)}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{c + a - b + a + b - c}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies \frac{2 a}{{\left(b + c\right)}^{2} - {a}^{2}} = \frac{2 a}{{a}^{2} - {\left(b - c\right)}^{2}}$

$\implies {\left(b + c\right)}^{2} - {a}^{2} = {a}^{2} - {\left(b - c\right)}^{2}$

$\implies {\left(b + c\right)}^{2} + {\left(b - c\right)}^{2} = {a}^{2} + {a}^{2}$

$\implies 2 {b}^{2} + 2 {c}^{2} = 2 {a}^{2}$

$\implies {b}^{2} + {c}^{2} = {a}^{2}$

This means the triangle is right angled.

3

Prove  (1+sinβ-cosβ)/(1+sinβ+cosβ) + (1+sinβ+cosβ)/(1+sinβ-cosβ) = 2cosecβ?

dk_ch
Featured yesterday

1st part of LHS

 =(1+sinβ-cosβ)/(1+sinβ+cosβ)

 =(sinbeta(1+sinβ-cosβ))/(sinbeta(1+sinβ+cosβ))

 =(sinbeta+sin^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

 =(sinbeta+1-cos^2β-sinbetacosβ)/(sinbeta(1+sinβ+cosβ))

 =(sinbeta(1-cosbeta)+(1-cosβ)(1+cosβ))/(sinbeta(1+sinβ+cosβ))

 =((1-cosbeta)(sinbeta+1+cosβ))/(sinbeta(1+sinβ+cosβ))

$= \frac{1 - \cos \beta}{\sin} \beta$

$= \frac{1}{\sin} \beta - \cos \frac{\beta}{\sin} \beta$

$= \csc \beta - \cot \beta$

2nd part of LHS
=(1+sinβ+cosβ)/(1+sinβ-cosβ)

 =(sinbeta(1+sinβ+cosβ))/(sinbeta(1+sinβ-cosβ))

 =(sinbeta+sin^2β+sinbetacosβ)/(sinbeta(1+sinβ-cosβ))

 =(sinbeta+1-cos^2β+sinbetacosβ)/(sinbeta(1+sinβ-cosβ))

 =(sinbeta(1+cosbeta)+(1-cosβ)(1+cosβ))/(sinbeta(1+sinβ-cosβ))

 =((1+cosbeta)(sinbeta+1-cosβ))/(sinbeta(1+sinβ-cosβ))

$= \frac{1 + \cos \beta}{\sin} \beta$

$= \frac{1}{\sin} \beta + \cos \frac{\beta}{\sin} \beta$

$= \csc \beta + \cot \beta$

LHS
 = cscβ-cotbeta+cscbeta+cotbeta

$= 2 \csc \beta = R H S$

1

If A = <3 ,8 ,-1 >, B = <4 ,-3 ,-1 > and C=A-B, what is the angle between A and C?

Somebody N.
Featured yesterday

${26.54}^{\circ}$

Explanation:

$A = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right]$

$B = \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right]$

$C = A - B = \left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] - \left[\begin{matrix}4 \\ - 3 \\ - 1\end{matrix}\right] = \left[\begin{matrix}3 - 4 \\ 8 - \left(- 3\right) \\ - 1 - \left(- 1\right)\end{matrix}\right] = \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right]$

We can find the angle between vectors using the Dot Product

The dot product states that for vectors a and b:

$a \cdot b = | | a | | \cdot | | b | | \cdot \cos \left(\theta\right)$

The dot product is sometimes called the inner product, because of the way the vectors a and b are multiplied and summed.

In algebra we are used to multiplying brackets in the following way.

$\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$

In the case of the dot product we multiply the vectors in the following way.

$\left(a + b + c\right) \cdot \left(d + e + f\right) = a d + b e + c f$

So we are just multiplying corresponding components and then adding them together.

Let $a = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right]$

Magnitude of $a = | | a | |$

$| | a | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$

Now to our example:

First find the product of:

$A \cdot C$

$\left[\begin{matrix}3 \\ 8 \\ - 1\end{matrix}\right] \cdot \left[\begin{matrix}- 1 \\ 11 \\ 0\end{matrix}\right] = \left[\begin{matrix}3 \times - 1 \\ 8 \times 11 \\ - 1 \times 0\end{matrix}\right]$

$= \left[\begin{matrix}- 3 \\ 88 \\ 0\end{matrix}\right] = - 3 + 88 + 0 = 85$

Now find the magnitudes of A and C:

$| | A | | = \sqrt{{\left(3\right)}^{2} + {\left(8\right)}^{2} + {\left(- 1\right)}^{2}} = \sqrt{74}$

$| | C | | = \sqrt{{\left(- 1\right)}^{2} + {\left(11\right)}^{2} + {\left(0\right)}^{2}} = \sqrt{122}$

So we have for:

$a \cdot b = | | a | | \cdot | | b | | \cdot \cos \left(\theta\right)$

$85 = \sqrt{74} \cdot \sqrt{122} \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{85}{\sqrt{74} \cdot \sqrt{122}}$

$\theta = \arccos \left(\cos \left(\theta\right)\right) = \arccos \left(\frac{85}{\sqrt{74} \cdot \sqrt{122}}\right) = {26.54}^{\circ}$
( 2 .d.p.)

So the angle between vectors A and C is ${26.54}^{\circ}$

Notice from the diagram that the angle found by the dot product, is the angle between the vectors where they are pointing in the same relative direction.