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## Please Help! Can anyone help me with either part a or part b? I'm completely confused :( How do you find alpha and beta? I only have to answer either part a or part b so whichever would be very helpful!!

dk_ch
Featured 3 months ago

Part(b)

Let hypotenuse of the lower triangle be h . Then the opposite of ${25}^{\circ}$ will be $= h \sin {25}^{\circ}$ and the adjacent will be $= h \cos {25}^{\circ}$

For upper triangle
The hypotenuse $= h \sec \alpha$
and opposite to $\alpha$ is $= h \tan \alpha$

By the condition of part(b) of the given problem the perimeter of both the triangle (upper and lower) are same. So we can say that the sum of other two sides excluding common side will be same for both the triangles.

Hence

$h \sec \alpha + h \tan \alpha = h \cos {25}^{\circ} + h \sin {25}^{\circ}$

$\implies \sec \alpha + \tan \alpha = \cos {25}^{\circ} + \sin {25}^{\circ} = 1.3289 \ldots \left(1\right)$

Now

$1.3289 \left(\sec \alpha - \tan \alpha\right) = {\sec}^{2} \alpha - {\tan}^{2} \alpha = 1$

$\implies \sec \alpha - \tan \alpha = \frac{1}{1.3289}$

$\implies \sec \alpha - \tan \alpha = 0.7525 \ldots \left(2\right)$

Adding (1) and (2) we get

$2 \sec \alpha = 2.0814$

$\implies \sec \alpha = 1.0407$

$= \cos \alpha = 0.9609$

$\implies \alpha = {\cos}^{-} 1 \left(0.9609\right) = {16.077}^{\circ}$

And

$\beta = {90}^{\circ} - \alpha = {73.923}^{\circ}$

Part-(a)

By the given condition of part (a) of the question

Area of the upper triangle = Area of the lower triangle

$\implies \frac{1}{2} \times h \times h \tan \alpha = \frac{1}{2} h \cos {25}^{\circ} \times h \sin {25}^{\circ}$

$\tan \alpha = \frac{1}{2} \times 2 \cos {25}^{\circ} \times \sin {25}^{\circ} = \frac{1}{2} \times \sin {50}^{\circ} = 0.3830$

$\implies \alpha = {\tan}^{-} 1 \left(0.3830\right) = {20.958}^{\circ}$

Then $\beta = {90}^{\circ} - {20.95}^{\circ} = {69.042}^{\circ}$

## An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(2 ,1 )# to #(3 ,7 )# and the triangle's area is #12 #, what are the possible coordinates of the triangle's third corner?

Tom G.
Featured 3 months ago

The coordinates of C are either ($- 1.3 \dot{91} \dot{8} , 4. \dot{64} \dot{8}$) or ($6.3 \dot{91} \dot{8} , 3. \dot{35} \dot{1}$)

This question took a lot longer to answer than I first anticipated!

#### Explanation:

There are two possible solutions for point C, both of which must lie on the line bisecting AB, as ABC is isosceles and all the points on this perpendicular line are equidistant from points A and B.

The mid-point of AB is straightforward enough:

$\left(\frac{2 + 3}{2} , \frac{1 + 7}{2}\right) = \left(\frac{5}{2} , \frac{8}{2}\right) = \left(2.5 , 4\right)$

AB has a slope of 6, because an increase of 1 on the $x$ axis results in an increase of 6 on the $y$ axis, and the slope of the perpendicular on which ${C}_{1}$ and ${C}_{2}$ lie has a slope of -1/6, because an increase of 6 along the $x$ axis results in a decrease of 1 on the $y$ axis.

Now triangle ABC has an area of 12

The area $A$ of a triangle is defined as $A = \frac{1}{2} b h$, where $b$ is the length of the base and $h$ is the height. In this case, we can consider side AB to be the base $b$, and the distance between the mid-point of AB and C to be the height $h$.

AB = $\sqrt{{\left(7 - 1\right)}^{2} + {\left(3 - 2\right)}^{2}} = \sqrt{{6}^{2} + {1}^{2}} = \sqrt{37} = b$

So with this value for $b$ we can solve for $h$ as follows:

$12 = \frac{1}{2} \times \sqrt{37} \times h$

$\therefore h = \frac{24}{\sqrt{37}}$

So point C must lie on one side or the other of AB, a distance of $\frac{24}{\sqrt{37}}$ along the line which bisects it.

In order to calculate the coordinates, we can imagine a right-angled triangle with a hypotenuse of length $\frac{24}{\sqrt{37}}$ with the opposite side length $n$ and the adjacent side length $6 n$

From Pythagorus, we know that ${h}^{2} = {a}^{2} + {o}^{2}$

$\therefore {\left(\frac{24}{\sqrt{37}}\right)}^{2} = 36 {n}^{2} + {n}^{2} = 37 {n}^{2}$

$\therefore {n}^{2} = {\left(\frac{24}{\sqrt{37}}\right)}^{2} / 37$

So $n = \frac{24}{37} = o$ and $6 n = \frac{144}{37} = a$

Point ${C}_{1}$ on the left of AB will have coordinates as follows:

$\left(2.5 - a , 4 + o\right)$

$2.5 - \frac{144}{37} = - 1.3 \dot{91} \dot{8}$

$4 + \frac{24}{37} = 4. \dot{64} \dot{8}$

So ${C}_{1} = \left(- 1.3 \dot{91} \dot{8} , 4. \dot{64} \dot{8}\right)$

Point ${C}_{2}$ on the right of AB will have coordinates as follows:

$\left(2.5 + a , 4 - o\right)$

$2.5 + \frac{144}{37} = 6.3 \dot{91} \dot{8}$

$4 - \frac{24}{37} = 3. \dot{35} \dot{1}$

So ${C}_{2} = \left(6.3 \dot{91} \dot{8} , 3. \dot{35} \dot{1}\right)$

## A person walks around an equilateral triangle with side length of 4 ft long If the person maintains a distance of 4 ft. from the triangle at all times how far must the person walk around fully?

Alan P.
Featured 1 month ago

$12 + 8 \pi$ feet

#### Explanation:

Walking a straight line parallel to each side, the person will walk $3 \times 4 \text{ feet}$

At each corner the person must walk $\frac{1}{3}$ of a circle with radius $4 \text{ feet}$ to maintain his/her distance of $4 \text{ feet}$ from the triangle:
$3 \times \left(\frac{1}{3} \cdot 2 \cdot 4 \cdot \pi\right) = 8 \pi \text{ feet}$

For a total distance of $12 + 8 \pi \text{ feet}$

## Triangle ABC has AB=10, BC=14, and AC=16. What is the perimeter of triangle DEF created by each vertex being the midpoint of AB, BC and AC?

CW
Featured 1 month ago

$20$

#### Explanation:

Given $A B = 10 , B C = 14 \mathmr{and} A C = 16$,

Let $D , E \mathmr{and} F$ be the midpoint of$A B , B C \mathmr{and} A C$, respectively.

In a triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length.

$\implies D E$ is parallel to $A C , \mathmr{and} D E = \frac{1}{2} A C = 8$
Similarly, $D F$ is parallel to $B C , \mathmr{and} D F = \frac{1}{2} B C = 7$
Similarly, $E F$ is parallel to $A B , \mathmr{and} E F = \frac{1}{2} A B = 5$

Hence, perimeter of $\Delta D E F = 8 + 7 + 5 = 20$

side note : $D E , E F \mathmr{and} F D$ divide $\Delta A B C$ into 4 congruent triangles, namely, $\Delta D B E , \Delta A D F , \Delta F E C \mathmr{and} \Delta E F D$

These 4 congruent triangles are similar to $\Delta A B C$

## Three vertices of parallelogram WXYZ are W(3,1), X(2,7),and Z(4,0). How do I find the coordinates of vertex Y?

CW
Featured 2 weeks ago

$\left(3 , 6\right) \mathmr{and} \left(1 , 8\right)$

#### Explanation:

Some of the properties of a parallelogram:
a) Opposite sides are parallel.
b) Opposite sides are congruent.
c) Opposite angles are congruent.
d) The diagonals bisect each other, $\implies$ the two diagonals have the same midpoint.

The Midpoint Formula: The midpoint of two points, $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , y 2\right)$ is the point M found by the following formula :

$M = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

First, let $X Z \mathmr{and} W Y$ be the two diagonals,
$\implies X Z \mathmr{and} W Y$ have the same midpoint,

$\implies \frac{2 + 4}{2} = \frac{3 + x}{2} , \implies x = 3$
$\implies \frac{7 + 0}{2} = \frac{1 + y}{2} , \implies y = 6$

Hence, the first coordinates of vertex $Y$ are $\left(3 , 6\right)$

Then, let $X W \mathmr{and} Y Z$ be the two diagonals,
$\implies X W \mathmr{and} Y Z$ have the same midpoint.

$\implies \frac{2 + 3}{2} = \frac{x + 4}{2} , \implies x = 1$
$\implies \frac{7 + 1}{2} = \frac{y + 0}{2} , \implies y = 8$

So the second coordinates for vertex Y are $\left(1 , 8\right)$

Hence, the two possible coordinates for vertex Y are $\left(3 , 6\right) \mathmr{and} \left(1 , 8\right)$

## Line #bar(AB)# has an equation of line given by #y_(AB)=1/2x#. The line is intersected by another line, #bar(BC)# at point B(1,1/2) forming and angle of #pi/6#. What is the equation of line of #bar(BC)#? "Hint: you can use vector algebra"

Shwetank Mauria
Featured 9 hours ago

Equation of $B C$ is
either $\left(4 \sqrt{3} - 2\right) y - \left(4 + 2 \sqrt{3}\right) x + 5 = 0$
or $\left(4 \sqrt{3} + 2\right) y + \left(4 - 2 \sqrt{3}\right) x - 5 = 0$

#### Explanation:

The slope of the line $A B$ given by $y = \frac{1}{2} x$ is $\frac{1}{2}$

Let the slope of other line be $m$ i.e. it is of the type $y = m x + c$. But as it passes through $\left(1 , \frac{1}{2}\right)$, we have $\frac{1}{2} = m + c$ or $c = \frac{1}{2} - m$.

As the slope of two lines is $\frac{1}{2}$ and $m$, the angle between them will be given by

$\tan \left(\pm A\right) = \frac{m - \frac{1}{2}}{1 + m \times \frac{1}{2}} = \frac{2 m - 1}{2 + m}$

As angle is $\frac{\pi}{6}$ and $\tan \left(\frac{\pi}{6}\right) = \pm \frac{1}{\sqrt{3}}$ and hence

either $\frac{2 m - 1}{2 + m} = \frac{1}{\sqrt{3}}$ i.e. $2 \sqrt{3} m - \sqrt{3} = 2 + m$ and $m = \frac{2 + \sqrt{3}}{2 \sqrt{3} - 1}$

and then $c = \frac{1}{2} - \frac{2 + \sqrt{3}}{2 \sqrt{3} - 1} = \frac{2 \sqrt{3} - 1 - 4 - 2 \sqrt{3}}{4 \sqrt{3} - 2} = - \frac{5}{4 \sqrt{3} - 2}$

and equation of line is $y = \frac{2 + \sqrt{3}}{2 \sqrt{3} - 1} x - \frac{5}{4 \sqrt{3} - 2}$

or $\left(4 \sqrt{3} - 2\right) y - \left(4 + 2 \sqrt{3}\right) x + 5 = 0$

or $\frac{2 m - 1}{2 + m} = - \frac{1}{\sqrt{3}}$ i.e. $2 \sqrt{3} m - \sqrt{3} = - m - 2$ and $m = - \frac{2 - \sqrt{3}}{2 \sqrt{3} + 1}$

and then $c = \frac{1}{2} + \frac{2 - \sqrt{3}}{2 \sqrt{3} + 1} = \frac{2 \sqrt{3} + 1 + 4 - 2 \sqrt{3}}{4 \sqrt{3} + 2} = \frac{5}{4 \sqrt{3} + 2}$

and equation of line is $y = - \frac{2 - \sqrt{3}}{2 \sqrt{3} + 1} x + \frac{5}{4 \sqrt{3} + 2}$

or $\left(4 \sqrt{3} + 2\right) y + \left(4 - 2 \sqrt{3}\right) x - 5 = 0$
$\left(\left(4 \sqrt{3} - 2\right) y - \left(4 + 2 \sqrt{3}\right) x + 5\right) \left(\left(4 \sqrt{3} + 2\right) y + \left(4 - 2 \sqrt{3}\right) x - 5\right) \left(2 y - x\right) = 0$
graph{((4sqrt3-2)y-(4+2sqrt3)x+5)((4sqrt3+2)y+(4-2sqrt3)x-5)(2y-x)=0 [-4, 4, -2, 2]}

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