# Math Help - Dreaded Word Problems

So I was doing my homework quite merrily until I go to the end where I ran into word problems.. (why do they always stick them at the end)

Can you guys point me into the right direction?

Q#1.

The minute hand on a clock is 6 cm long and the hour hand is 4 cm long. Give an expression for d , the distance between the tips of the hands, in terms of a , the angle between the hands.

What Does "d"=?

My Plan of Attack: I drew a lame picture of the clock hands neatly labeled. I noticed you can form a triangle and thus use Pythagorean to find the distance between the 2 points. But Im not sure which side is the longer side? Do you guys think it should be $6^2+4^2=c^2?$

2nd, I think, once I know the 3 side lengths, I can use either Cosine law or Sine law to find the angle..

And what is the question asking when they are asking for an expression for "d"?

Q#2.

At noon, ship A passes 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 5 km/h. Give an expression for d , the distance between the two ships, in terms of t , the number of hours after noon.

My Plan of Attack: Confused.

I have no idea why I can't handle word problems.

TIA as always

2. Originally Posted by mvho
So I was doing my homework quite merrily until I go to the end where I ran into word problems.. (why do they always stick them at the end)

Can you guys point me into the right direction?

Q#1.

The minute hand on a clock is 6 cm long and the hour hand is 4 cm long. Give an expression for d , the distance between the tips of the hands, in terms of a , the angle between the hands.

What Does "d"=?

My Plan of Attack: I drew a lame picture of the clock hands neatly labeled. I noticed you can form a triangle and thus use Pythagorean to find the distance between the 2 points. But Im not sure which side is the longer side? Do you guys think it should be $6^2+4^2=c^2?$

2nd, I think, once I know the 3 side lengths, I can use either Cosine law or Sine law to find the angle..

And what is the question asking when they are asking for an expression for "d"?
Join the tips of the minute hand and hour hand,
You have a triangle: 1 side is 6cm long, second side is 4cm long and third side is d cm.
Since you know the angle between the known sides:Apply cosine law.

3. Yes, That is what I thought but I was more confused at what they actually want me to find?

And does it matter if the unknown side is the longer side or shorter side? I am not clear about that as well.

Do you guys think it should be $6^2+4^2=c^2 OR, c^2+4^2=6^2?$

4. Originally Posted by mvho

Q#2.

At noon, ship A passes 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 5 km/h. Give an expression for d , the distance between the two ships, in terms of t , the number of hours after noon.
See figure.

Ship A started from A at noon. After time t, it is at C
hence, AC=10t

Ship B started at harbour(B) at 9am. At noon, it was at D.
BD=????
At time t after noon, it is at E.
DE=????

5. Originally Posted by mvho
Yes, That is what I thought but I was more confused at what they actually want me to find?

And does it matter if the unknown side is the longer side or shorter side? I am not clear about that as well.

Do you guys think it should be $6^2+4^2=c^2 OR, c^2+4^2=6^2?$
Neither! You only use the Pythagorean Theorem for right triangles, and for the most part, the triangles formed by the two hands plus the distance between them at the tips are not right triangles. You have a triangle where you know two of the sides, and the angle between them is a. The unknown side is d. As malaygoel said, just use the Law of Cosines:

\begin{aligned}
d^2 &= 4^2 + 6^2 - 2(4)(6)\cos a \\
d^2 &= 52 - 48\cos a \\
d &= \sqrt{52 - 48\cos a}
\end{aligned}

See diagram below.

01

6. Thanks everyone

Ship A started from A at noon. After time t, it is at C
hence, AC=10t

Ship B started at harbour(B) at 9am. At noon, it was at D.
BD=????
At time t after noon, it is at E.
DE=????
So, BD=5t
and AC=10t

Is it also true that the length from point A to point B is 8km since the question said it passes 8k west? If this is the case, I can use pythagorean since a rectangle would be a right triangle.

I think I should have:

$d=\sqrt{8^2+5t^2}$

Since I know that AC=5t and BD=5t so if I minus 10t-5t, I'll know the length.