January 11, 2008
PHY2053 Discussion
Warm Up Quiz (Appendix A & Chapter 1.7, 1.8)
Name:
UFID:
**1. (5 pts) A tower having a height of 100. m is located near a building. An observer
whose eyes are 1.60 m high is on the top of the building. He/she observes the top of the
tower at an angle of 45.0º above the horizontal and the base of the tower at an angle
30.0º below the horizontal. What is the height of the building the observer is on?
Let x be the distance between the building and the tower, y be the height of the part of
the tower form ground to the observer’s eye level, and y’ be the height of the rest of the
tower.
Using the definition of tanθ, we get
tan30º = y/x (1) and tan45º = y’/x (2)
Since y+y’ = 100, we get
100 = x(tan30º+tan45º) ⇒ x = 100/(tan30º+tan45º) = 63.4m
Plugging the result into (1), we get
y = xtan30º = 63.4tan30º = 36.6m
Thus the height of the building is given by
h = 36.6-1.6 = 35.0m
*2. (5 pts) A right triangle is uniquely determined if you specify a hypotenuse and one of
the other sides. Now consider a triangle having a hypotenuse of length 6 cm and one of
the other sides 3 cm. Determine the unknown side and angles and draw the triangle.
Let the sides be a(=6cm), b(=3cm) and c, and the corresponding angles be A(=90º), B and
C.
Using Pythagorean theorem and the definition of sinθ and cosθ, we get
c = √(6²-3²) = √27 = 3√3 cm(=5.20cm)
sinB = 3/6 ⇒ B = sinˉ¹(3/6) = 30º, cosC = 3/6 ⇒ C = cosˉ¹(3/6) = 60º.
You can get this right triangle by folding an equilateral triangle at the center line.
***3. (5 pts) Except for right triangles (see the previous problem), normally a triangle is
not uniquely determined when you specify two sides and an angle opposite to one of
them. This time, we consider triangles having sides of length 6 cm and 2√3 cm. The
angle opposite to 2√3cm side is 30º. Determine the unknown sides and angles of all the
triangles satisfy this condition and draw them.
Let the sides be a(=6cm), b(=2√3cm) and c, and the corresponding angles be A, B(=30º)
and C. Using law of cosines, we get
b² = a²+c²-2accos30º ⇒ (2√3)² = 6²+c²-2×6c(√3/2) ⇒ c²-6√3+24 = 0 (1)
Solve (1) by using the formula for quadratic equations (See [A.8] in Appendix A.):
c = [6√3±√(108-96)]/2 = (6√3±2√3)/2 = 4√3, 2√3
When c = 2√3, it is an isosceles. (C= 30º, A = 180-30-30 = 120º)
When c = 4√3, using law of cosines, we get
c² = a²+b²-2abcosC ⇒ cosC = (a²+b²-c²)/(2ab) = [6²+(2√3)²-(4√3)²]/(2×6×2√3) = 0.
Thus C = cosˉ¹(0) = 90º.
It is a right triangle. (A = 180-30-90 = 60º)
**4. (5 pts) Solve the following equation for x: 3^(2x+1)+5×3^(x)-2 = 0. (2x+1 and x are
the powers to which 3 is raised.)
Using rules of exponents, we get
3(3^x)²+5(3^x)-2 = 0
We solve this quadratic equation by factoring (or you can just use the formula):
[3(3^x)-1][(3^x)+2] = 0 ⇒3(3^x)-1=0 or (3^x)+2 = 0.
Since 3^x must be positive,
3^x = 1/3
Thus x = -1.