Bunch of questions that may have typographical errors in the answer key

[BSElectrician]

I have a problem that involves these questions, what i need you to do is to verify if the textbook answers are typographical errors. Ill show my solutions and have to tell me which part i get wrong.

Homework Statement

1. A corner lot is 35m on one street and 25m on the other street. The angle between the two lines of the street being 82 degrees. The other to two lines of the lot are respectively perpendicular to the lines of the streets. What is the worth of the lot if its unit price is $2500 per square m?

textbook Ans. 1,884,050

2. Find the distance between (5, 30degrees) and (-8, 50 degrees).

Ans. 10.14

3. The angle from line 1 to line 2 is arctan(2/3) and the slope of line 1 is -1. Find the slope of line 2.

Ans. -1/5

4. the sides of a cyclic quadrilateral measures 8cm, 9cm, 12cm and 7cm. Find the area of the circumscribing circle

Ans. 134.37 cm^2

5. How far from a vertex is the opposite face of a tetrahedron if an edge is 50cm long?


Ans: 40.825cm

Homework Equations

The Attempt at a Solution

1. I will subdivide the total lot into 2 triangles so i can calculate the area

[itex]A_{1} = \frac{1}{2}absin{\theta} [/itex]

[itex]A_{1} = \frac{1}{2}(25)(35)sin(82) [/itex]

[itex]A_{1} = 433.24[/itex]

Law of cosine for the opposite side: o = opposite side

[itex]o = \sqrt{35^2 + 25^2 - 2(25)(35)\cos{82}}[/itex]

[itex]o = 40.1[/itex]

opposite angle of 82 degrees
360 - 90 - 90 - 82 = 98 degrees

angles of the second triangle are 98/2 = 41 since they have both 90 degrees angle
hence they also have the congruent sides

law of sine to find distance d:
[itex]\frac{\sin(98)}{40.1} = \frac{\sin(41)}{d}[/itex]

[itex]d = 26.5665[/itex]

[itex]A_{2} = \frac{1}{2}(26.5665)^{2}\sin{98} [/itex]

[itex]A_{2} = 349.455[/itex]

[itex]A_{T} = A_{1} + A_{2} = 782.695 sq.m [/itex]

[itex]Cost_{total} = 782.695(2500) = 1,956,743.613[/itex]

Ans. 1,956,743.613 dollars

2.
5 in the 1st quadrant
8 in the 3rd quadrant

Angle between the two sides = 90 + 30 + 40 = 160
Law of cosine

[itex]d^2 = 5^2 + 8^2 -2(5)(8)\cos(160)[/itex]
[itex]d = 12.81[/itex]

Ans. 12.81

3.
Formula
[itex]\theta = \frac{m_{2} - m_{1}}{1+m_{2}m_{1}}[/itex]
since:
[itex]\tan^{-1}(\frac{2}{3}) = \frac{m_{2} - (-1)}{1+m_{2}(-1)}[/itex]

Ans. [itex]m_{2} = -0.26[/itex]

4.
Area of cyclic Quadrilateral
[itex]A_{cyc} = \sqrt{(s-a)(s-b)(s-c)(s-d)}[/itex]
where
[itex]s = \frac{a + b + c + d}{2} [/itex]

and the radius of the circumscribing circel
[itex]r = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A_{cyc}}[/itex]

[itex]s = \frac{9 + 8 + 7 + 12}{2} [/itex]

[itex]s = 18 [/itex]

[itex]A_{cyc} = \sqrt{ (18-9)(18-8)(18-7)(18-12) }[/itex]

[itex]A_{cyc} = 69.714[/itex]

[itex]r = \frac{\sqrt{((9)(8)+(7)(12))(9(7)+8(12))(9(12)+8(7))}}{4(69.714)}[/itex]

[itex]r = 7.233[/itex]

[itex]A_{circ} = \pi{7.233}^2[/itex]

Ans. [itex]A_{circ} = 164.37[/itex]

5.
A tetrahedron is a triangular shaped object with equal sides
So i will assume that the shortest distance will be from the central side of the face to the angle bisector opposite to that face.

Then on the side view of the tetrahedron we have a equilateral triangle
i cut the triangle in half so that halved triangle has angles 30,60,90 and d is distance from vertex to face
[itex]\sin(\theta) = \frac{opposite}{hypotenuse}[/itex]

[itex]\sin(60) = \frac{d}{50} [/itex]

Ans. d = 43.30cm

 

[NascentOxygen]

Q1) $1,833,997
Q2) 12.8131
Q3) -0.2
m₁ is -45 deg, diff is 33.69 deg, so m₂ is -11.31 deg.
tan -11.31 is -0.2

 

[SammyS]

...
1. A corner lot is 35m on one street and 25m on the other street. The angle between the two lines of the street being 82 degrees. The other to two lines of the lot are respectively perpendicular to the lines of the streets. What is the worth of the lot if its unit price is $2500 per square m?

textbook Ans. 1,884,050

The Attempt at a Solution

1. I will subdivide the total lot into 2 triangles so i can calculate the area

[itex]A_{1} = \frac{1}{2}absin{\theta} [/itex]

[itex]A_{1} = \frac{1}{2}(25)(35)sin(82) [/itex]

[itex]A_{1} = 433.24[/itex]

Law of cosine for the opposite side: o = opposite side

[itex]o = \sqrt{35^2 + 25^2 - 2(25)(35)\cos{82}}[/itex]

[itex]o = 40.1[/itex]

opposite angle of 82 degrees
360 - 90 - 90 - 82 = 98 degrees

angles of the second triangle are 98/2 = 41 since they have both 90 degrees angle
hence they also have the congruent sides

...

That should be 180° - 98° = 82°, and 82°/2 = 41°. However, those two angles are not congruent !

This is clear because the two unknown angles of the first triangle can't possibly be congruent. Use the law of sines to find these two angles.

Spoiler

38.1465° and 59.8535° in your first triangle.

 

[NascentOxygen]

Q4) 134.4

 

[Joffan]

Q5) 40.82483cm
(First draw the triangle that cuts the tetrahedron in half by cleaving one edge and bisecting two faces, then drop a perpendicular from a vertex).