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

1. Dec 27, 2011

### 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.

1. The problem statement, all variables and given/known data

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 2. Relevant equations 3. The attempt at a solution 1. I will subdivide the total lot into 2 triangles so i can calculate the area $A_{1} = \frac{1}{2}absin{\theta}$ $A_{1} = \frac{1}{2}(25)(35)sin(82)$ $A_{1} = 433.24$ Law of cosine for the opposite side: o = opposite side $o = \sqrt{35^2 + 25^2 - 2(25)(35)\cos{82}}$ $o = 40.1$ 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: $\frac{\sin(98)}{40.1} = \frac{\sin(41)}{d}$ $d = 26.5665$ $A_{2} = \frac{1}{2}(26.5665)^{2}\sin{98}$ $A_{2} = 349.455$ $A_{T} = A_{1} + A_{2} = 782.695 sq.m$ $Cost_{total} = 782.695(2500) = 1,956,743.613$ 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 $d^2 = 5^2 + 8^2 -2(5)(8)\cos(160)$ $d = 12.81$ Ans. 12.81 3. Formula $\theta = \frac{m_{2} - m_{1}}{1+m_{2}m_{1}}$ since: $\tan^{-1}(\frac{2}{3}) = \frac{m_{2} - (-1)}{1+m_{2}(-1)}$ Ans. $m_{2} = -0.26$ 4. Area of cyclic Quadrilateral $A_{cyc} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s = \frac{a + b + c + d}{2}$ and the radius of the circumscribing circel $r = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A_{cyc}}$ $s = \frac{9 + 8 + 7 + 12}{2}$ $s = 18$ $A_{cyc} = \sqrt{ (18-9)(18-8)(18-7)(18-12) }$ $A_{cyc} = 69.714$ $r = \frac{\sqrt{((9)(8)+(7)(12))(9(7)+8(12))(9(12)+8(7))}}{4(69.714)}$ $r = 7.233$ $A_{circ} = \pi{7.233}^2$ Ans. $A_{circ} = 164.37$ 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 $\sin(\theta) = \frac{opposite}{hypotenuse}$ $\sin(60) = \frac{d}{50}$ Ans. d = 43.30cm Last edited: Dec 27, 2011 2. Dec 27, 2011 ### NascentOxygen ### Staff: Mentor 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

3. Dec 27, 2011

### SammyS

Staff Emeritus
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.

38.1465° and 59.8535° in your first triangle.

4. Dec 28, 2011

### Staff: Mentor

Q4) 134.4

5. Dec 30, 2011

### 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).

Last edited: Dec 30, 2011