# Trigonometry Question... (Possibly Using the Cosine Rule)

• April 9th 2010, 11:05 PM
Trigonometry Question... (Possibly Using the Cosine Rule)
Hello,

In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed... (Headbang)

An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?

Many thanks!
• April 10th 2010, 12:01 AM
sa-ri-ga-ma
Quote:

Hello,

In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed... (Headbang)

An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?

Many thanks!

If the foot of the perpendicular from the balloon is x m from left end, then
x^2 + h^2 = 120^2.........(1)
(35 - x)^2 + h^2 = 100^2 .....(2)
solve these two equations and find x. Substitute this value in eq. 1 to find h.
• April 10th 2010, 05:25 PM
Hmmm...
As I am new to simultaneous equations, please correct me if I am wrong:

x^2 + h^2 = 120^2 ....... (1)
(35 - x)^2 + h^2 = 100^2 ....... (2)

x^2 + h^2 = 120^2 ....... (1)
35^2 -70x + x^2 + h^2 = 100^2 ....... (2)

How do I proceed? Adding or subtracting x^2 or h^2 ?
• April 10th 2010, 06:24 PM
sa-ri-ga-ma
Quote:

As I am new to simultaneous equations, please correct me if I am wrong:

x^2 + h^2 = 120^2 ....... (1)
(35 - x)^2 + h^2 = 100^2 ....... (2)

x^2 + h^2 = 120^2 ....... (1)
35^2 -70x + x^2 + h^2 = 100^2 ....... (2)

How do I proceed? Adding or subtracting x^2 or h^2 ?

From eq.1 you get
h^2 = 120^2 - x^2
Put it in the second equation and solve for x.
• April 10th 2010, 06:34 PM
So this is, indeed, a simultaneous equation?

Can I expand (35 - x) ^2? That is, I think, where I am going wrong.
• April 11th 2010, 02:30 AM
sa-ri-ga-ma
Quote:

So this is, indeed, a simultaneous equation?

Can I expand (35 - x) ^2? That is, I think, where I am going wrong.

In the eq.2 you have already expanded the above term.
35^2 -70x + x^2 + h^2 = 100^2 ....... (2)
Now substitute h^2 = 120^2 - x^2 in the above equation.
35^2 -70x + x^2 - x^2 + 120^2 = 100^2 ....... (2)
35^2 -70x + 120^2 = 100^2 ....... (2)
Now solve for x.
• April 11th 2010, 02:59 AM
Quote:

Hello,

In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed... (Headbang)

An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?

Many thanks!

To find the maximum height, using the cosine rule....

The height is maximum when the ropes are stretched to their limit,
forming a triangle.

Then, if we find the angle opposite the 100 m rope

$100^2=35^2+120^2-2(120)(35)cos\theta$

$2(120)35cos\theta=35^2+120^2-100^2$

$cos\theta=\frac{35^2+120^2-100^2}{(70)(120)}$

$\theta=arccos\left(\frac{35^2+120^2-100^2}{(70)(120)}\right)$

Then

$sin\theta=\frac{h}{120}\ \Rightarrow\ h=120sin\theta$