Download 11th quiz-solution

Bo Lin
11th Quiz - Math 32 Section 107 & 108
Solution
1. (4 pts) A rhombus is a convex quadrilateral whose all four sides have the
same length. If a rhombus has side length 2 and an angle π4 , find its area. (Hint:
a rhombus consists of two identical isosceles triangles)
Solution. If we connect one diagonal of the rhombus we get two identical isosceles triangles, each has two sides 2 and the angle between them π4 , then the area
√
√
of each triangle is 21 · 2 · 2 · sin π4 = 2 sin π4 = 2 22 = 2. So the area of the
√
√
rhombus is 2 · 2 = 2 2.
√
√
Partial credit: 2 sin π4 , 3 pts; 4 2 or 2, 2 pts; other wrong answers in
similar pattern, 1 pt.
Remark: This rhombus is NOT a regular 4-gon! Because its angles are not
all equal, actually the other two angles are π − π4 = 3π
4 .
2. (6 pts) In triangle ABC, suppose A = 80◦ , B = 70◦ and b = 4.
(1) Which side of the triangle is the longest?
(2) If sin 70◦ = d, find c. (d will appear in your answer)
Solution. (1)(3 pts) By the Law of Sines, the three sides are proportional to the
sine functions of the three angles. Since A = 80◦ , B = 70◦ , then C = 180◦ − A −
B = 30◦ . Note that y = sin x is increasing on [0, π2 ], then sin A > sin B > sin C.
¯ is the longest side.
So a (or BC)
Alternative proof: there is a famous result called ”larger(largest) angle facing
longer(longest) side”, which could be easily proved by the Law of Sines.
(2)(3 pts) By the Law of Sines, sinc C = sinb B . Note that b = 4, sin C =
sin 30◦ =
1
2
and sin B = sin 70◦ = d, then c =
1
2 ·4
d
= d2 .
◦
Partial credits: for (1), a without justification, 2 pts; b, 1 pt. For (2), 4 sind30 ,
2 pts;correctly applying the Law of Sines, 1 pt.
Remark: (for those who missed the section on Monday) the convention for
this part is to use upper case letter to denote vertices and angles of triangles,
and to use lower case letters to denote the length of the corresponding side. I
apologize for the confusion caused by d. In this situation d is neither an angle
nor a side length.
1
Bo Lin
3.
Suppose triangle ABC has side lengths 3, 5, 7 as follows:
(1) (4 pts) Find cos A;
(2) (2 pts) Find angle A;
(3) (4 pts) Find the area of triangle ABC.
Solution. (1) By condition a = 7, and the set {b, c} is {3, 5}. By the Law of
2
2
2
2
2
−72
1
= −15
Cosines, cos A = b +c2bc−a = 3 +5
2·3·5
30 = − 2 .
−1
(2) Since [0, π] is exactly the range of cos , A = cos−1 − 12 . While you
π
π
1
should know that cos π3 = 12 , so cos 2π
3 = cos(π − 3 ) = − cos 3 = − 2 . Then
2π
◦
A = 3 or 120 .
√
3
2π
(3) Since A = 2π
3 , then sin A = sin 3 = 2 . Then the area of triangle ABC
√
is
bc sin A
2
=
3·5·
2
3
2
=
√
15 3
4 .
2
2
2
Partial credit: (1), cos A = b +c2bc−a , 2 pts; incorrect version of it, 1 pt;
getting − 21 but answering −2 (which seems inconceivable), 3 pts. (2), ”correct”
π
−1
answer based on incorrect answer of (1), 2 pts; 4π
x where x
3 or 3 , 1 pt; cos
bc sin A
−1
is not in the domain of cos , 0 pt. (3)
, 2 pts; other correct formulae of
2
the area, 1 pt (because they are kind of useless in this situation).
Remark: Not distinguishing b and c is a huge hint, because then you are sure
that everything involved is symmetric between b, c. For example, if you switch
2
2
2
the roles of b and c, b +c2bc−a remains the same! This is exactly what symmetry
means.
I was shocked that even though I clarified on board, a few students still
labeled 3 to the longest side! I know the quiz could be stressful, while you
should always keeps common sense in your mind. Also maybe the position
of the upper case letter A is a little confusing, but you should know that it
represents the vertex instead of a side since it is upper case.
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