Help with this Math II Question?

Test Preparation SAT Subject Test Preparation
1

<p>What are all values of side a in the figure below such that two triangles can be constructed?
A. a > 4sqr root (3)
B. a > 8
C. a = 4sqr root (3)
D. 4sqr root (3) < a < 8
E. 8 < a < 8sqr root (3) </p>

<p>For the picture, see this link: triangle</a> | Flickr - Photo Sharing!</p>

2

<p>Are you sure there are no side lengths given? I feel like a piece of information is missing. Could be wrong though.</p>

3

<p>I’m pretty sure there is something missing. The question is dealing with law of Sines ambiguous case (SSA i think, but not sure) there should be one more side given.</p>

4

<p>The other two are right. For any triangle there’s always an infinite number of similar triangles, so there’s no way to do it without a side length.</p>

5

<p>sorry, there is.
the side next to the 60 degree angle shud say 8.
so here’s a redrawn picture: [url=&lt;a href=“http://www.■■■■■■■■■■/photos/63434575@N03/5770395918/]triangle”&gt;triangle 2 | aksgeh | Flickr]triangle</a> 2 | Flickr - Photo Sharing<img src=“sorry,%20my%20drawing%20skills%20aren’t%20very%20pro” alt=“/url”></p>

6

<p>The answer is D. This is a case of why is known as the “ambiguous case.” Think about it this way: if you drew the same triangle on a piece of paper and tried to draw different lengths of “a” with a compass, when would you get two sides? If a was less than 8, no triangle would be formed. If a was equal to the height of a triangle (8 sin 60), exactly one triangle would be formed. BUT, if it was greater than the height (8sin60=4 square root 3) and less than 8 (some length b), then the compass would cut the base two times, thus forming two separate triangles. All of this is based on the fact that sin theta=sine 180-theta. I hope that helped.</p>

7

<p>Ok, I understand where the 4 square roots of 3 comes from, but I don’t understand why it has to be less than 8. Can someone explain that? (sorry, but I don’t understand the drawing triangle thing…)</p>

8

<p>BUMPPP 10char</p>

9

<p>Just use the law of cosines. Let the unknown side be x.</p>

<p>a^2 = 8^2 + x^2 - 2<em>8</em>x*cos60 = 64 + x^2 - 8x</p>

<p>a^2 = x^2 - 8x + 64</p>

<p>Then go from there.</p>

10

<p>I understand where the 4 square root of 3 comes from, but I don’t get why it’s D instead of A. Why does the 8 play a role?
Also- another question.
A coin is tossed 3 times. Let A = {three heads occur} and B = {at least one head occurs} What is P (A U B)?
A) 1/8
B) 1/4
C) 1/2
D) 3/4
E) 7/8</p>

<p>Barrons says it is E. I say A. Who is right?</p>

11

<p>P(A U B) = P(A) + P(B) - P(A intersection B)</p>

<p>P(A) = P(Heads = 3) = 0.5^3 = 1/8
P(B) = P(Heads >=1) = 1 - (P(Heads = 0) + P(Heads = 1)) = 1 - 0.5^3 - 0.5*0.5^2
= 1 - 1/8 - 1/8 = 6/8</p>

<p>Now P(A intersection B) is 0 because A and B are independent events.</p>

<p>Sp P(A U B) = 6/8 + 1/8 = 7/8</p>

<p>Answer E)</p>

12

<p>a^2 = x^2 - 8x + 64</p>

<p>Solve this equation:</p>

<p>x^2 - 8x + 64 - a^2 = 0</p>

<p>x(1,2) = 8 ± root of(64 - 4<em>1</em>(64-a^2)) / 2</p>

<p>x(1,2) = 8± root of(4a^2 - 192) / 2</p>

<p>x(1,2) = 8 ± root of(4(a^2 - 48)) / 2</p>

<p>x(1,2) = 8 ±2*root of(a^2 - 48) / 2</p>

<p>x(1,2) = 4 ± root of(a^2 - 48)</p>

<p>x(1) = 4 + root of(a^2 - 48)
x(2) = 4 - root of(a^2 - 48)</p>

13

<p>If you plug in a=8, you get x(1) = 8 and x(2) = 0</p>

<p>If you plug in a=4*root(3), you get x(1)= 4 and x(2) = 4</p>

<p>So 4*root(3) < a< 8</p>

<p>Answer (d)</p>

14

<p>@jsanche32: Quick questions-

  1. On the probability problem, for the at least one part, why are you subtracting from 1 the probability of getting 1 head? Isn’t at least one just 1- probability of getting 0 heads? Also, why are the two dependent events? Won’t you just subtract out the probability of both getting 3 heads, because that is where they overlap?
  2. For the triangle question, so do you just plug it into Law of Cosines, get an expression, then plug in the answer choices?</p>
15

<p>2) correct</p>

<p>1) The question has two parts. One is p(heads = 3), and the
other is p(heads=> 1)</p>

<p>It’s not p(heads = 1) it’s the prob. of getting more than one head, which is p(heads=>1)</p>

<p>Which means it’s either; 1 head, 2 heads, or three heads</p>

<p>The prob . of that is: 1- [p(h=0) + p(h=1)]</p>

<p>Then, the two events are independent. i.e. They are separate. Throwing 3 heads. Then throwing three tails. Those are also independent events. The probability of one event has no effect on the probability of the other event,</p>

<p>In which case there is no intersection. P(A intersection B) = 0</p>

16

<p>I still don’t understand… I thought at least one was getting the probability of one head + probability of getting two heads + probability of getting three heads, or 1- probability of getting 0 heads.
So U means union right, which means probability of two events added minus the probability of both of them happening? I understand your explanation (i.e. the two aren’t contingent) but if you think of it in terms of a Venn diagram, it makes sense that getting 3 heads means that the two regions overlap… right?</p>

17

<p>EDIT: I get it now. Thanks.</p>