Hi There!

I had a maths exam today and there was this question......

Three horses are tethered at 3 corners of a triangular plot having sides 20m, 30m, 40m with ropes of 7m lengtth each. Find the area of this plot which can be grazed by the horses.

area of a sector = [p (radius)²q]/360 degrees.
Is there any way of solving this question???

Monalisa.

You could use the cosine rule. Label the sides a, b, c and the angle between sides a and b x. Then:

c² = a² + b² - 2ab cos(x)

Rearranging this should give the required value of x, but in fact you don't need to do this (as I realised while typing this) - use the fact that the interior angles of a triangle add up to 180 degrees . Together the horses can graze an area equivalent to a semicircle of radius 7.

Hi Tim!

Thanks for replying!

We are studying since grade 5 that sum of interior angles of a triangle is 180 degrees....but it just didn't come in my head at the time of my exam!!!

I just hope I won't be so careless the next time!

But would you mind telling me more about the cosine rule......we haven't studied it so far....but it's knowledge will definitely help!

Monalisa.

There are two useful trigonometric expressions which work even for non right-angled triangles. They are called the sine and cosine rules for reasons which will hopefully become obvious.

First consider an arbitrary triangle. Call the lengths of the sides a, b, c and label each angle by the capital letter corresponding to the side opposite it.

Draw a line through corner C perpendicular to side the side of length c. Call the length of this new line y.
[Diagram supplied by Brad Rodgers (P1930)]

This splits the triangle into two right-angled triangles, and we can use the ordinary trig expressions (which I'm assuming you know - tell me if you don't) to obtain:

y/b = sin A and y/a = sin B

Therefore b sin A = a sin B
and so b/(sin B) = a/(sin A)

This is the sine rule, and by repeating the same argument from a different corner you can see that this is also equal to c/(sin C)

Now call the length between corner A and the right angle x. Then by pythagoras:

y² = b² - x² = a² - ( c² -2cx +x² )

Also (by trig)

x/b = cos A

Therefore

a² = b² + c² - 2bc cos A

Which is the cosine rule. It is normally stated in the form I gave above, which is the same thing but with the triangle relabelled.

I hope some of that made sense.

Tim