# Thread: 3U Homework - Help!

1. ## 3U Homework - Help!

Hey guys, I need help with some Exam Revision - its really confusing...

a) The point M(-3,8) divides the interval joining A(-4,-9) and B(5,9) externally in the ratio k:1. Find the value of k.

b) Sydney Tower stands on a level ground, Three observers P,Q and R are observing Sydney tower from the ground. P is due north of Sydney Tower, R is due east of Uluru, and Q is on the line of sight from P to R and between them. The angles of elevation to the top of Sydney Tower from P, Q and R are 26 degrees, 28 degrees and 30 degrees, respectively. Find the bearing of Q from Sydney Tower.

c) The angle between the lines y = mx and y = 1/2 x is 45 degrees. Find the exact value of m.

I've tried, but nothing is working - If anyone could assist me in understanding these that would be terrific. Thank you for your time.

2. ## Re: 3U Homework - Help!

a) Use the formula

$x = \frac{mx_2+nx_1}{m+n}$ $y = \frac{my_2+ny_1}{m+n}$

Calculate the point $M$ using your ratio $k:-1$ and your two points (be careful, order matters).

Then compare your point to the one given and you should be able to find $k$.

c) Using the formula

$\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$

We get:

\begin{aligned} 1 &= \left|\frac{\frac 12-m}{1+\frac m2}\right| \\ \left|\frac 12-m\right| &= \left|1+\frac m2 \right|\end{aligned}

From here you can take cases, or you can use the definition $|x| = \sqrt{ x^2}$ and square both sides to solve for $m$.

Originally Posted by Heresy
Q is on the line of sight from P to R and between them.
Should that be read as "Q in the middle of P and R"? Or is it just saying the same thing twice?

3. ## Re: 3U Homework - Help!

Originally Posted by fan96
a) Use the formula

$x = \frac{mx_2+nx_1}{m+n}$ $y = \frac{my_2+ny_1}{m+n}$

Calculate the point $M$ using your ratio $k:-1$ and your two points (be careful, order matters).

Then compare your point to the one given and you should be able to find $k$.

c) Using the formula

$\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$

We get:

\begin{aligned} 1 &= \left|\frac{\frac 12-m}{1+\frac m2}\right| \\ \left|\frac 12-m\right| &= \left|1+\frac m2 \right|\end{aligned}

From here you can take cases, or you can use the definition $|x| = \sqrt{ x^2}$ and square both sides to solve for $m$.

Should that be read as "Q in the middle of P and R"? Or is it just saying the same thing twice?
My interpretation is that it is saying in the middle

4. ## Re: 3U Homework - Help!

Originally Posted by Heresy
My interpretation is that it is saying in the middle

Given that $QR = PQ$...

Let the height of Sydney Tower $OT$ be $h$.

(since the height of a tower won't change its bearing, we can use $h$ in our calculations but it will not appear in the final answer)

1. Find $OQ,\, OP,\,OR$ in terms of $h$.

2. Noting that $\triangle OPR$ is a right-angled, find $PR$ and hence $PQ$.

3. Then use the cosine rule to find $\angle POQ$.

5. ## Re: 3U Homework - Help!

Originally Posted by fan96

Given that $QR = PQ$...

Let the height of Sydney Tower $OT$ be $h$.

(since the height of a tower won't change its bearing, we can use $h$ in our calculations but it will not appear in the final answer)

1. Find $OQ,\, OP,\,OR$ in terms of $h$.

2. Noting that $\triangle OPR$ is a right-angled, find $PR$ and hence $PQ$.

3. Then use the cosine rule to find $\angle POQ$.
Mate you're a legend thank you so much, I really appreciate it!!

6. ## Re: 3U Homework - Help!

Originally Posted by fan96

Given that $QR = PQ$...

Let the height of Sydney Tower $OT$ be $h$.

(since the height of a tower won't change its bearing, we can use $h$ in our calculations but it will not appear in the final answer)

1. Find $OQ,\, OP,\,OR$ in terms of $h$.

2. Noting that $\triangle OPR$ is a right-angled, find $PR$ and hence $PQ$.

3. Then use the cosine rule to find $\angle POQ$.
I'm sorry to ask but can you help me with Part 2 of that solution process please? I'm utterly useless...

7. ## Re: 3U Homework - Help!

Originally Posted by Heresy
I'm sorry to ask but can you help me with Part 2 of that solution process please? I'm utterly useless...
Don't think negatively like that. Just keep practicing and you'll improve.

If you found $OR,\,OP,$ then use Pythagoras' theorem:

$OR^2+OP^2=PR^2$

and of course, $2PQ = PR$.

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