Calculate the height of the triangle to the nearest tenth

A triangular plot of land is enclosed by a fence. Two sides of the fence are 9.8 m and 6.6 m long, respectively. The other side forms an angle of 40 (degree) with the 9.8 m side.

(a) Calculate the height of the triangle to the nearest tenth. Compare it to the given sides.

(b) How many lengths are possible for the third side? Explain.

How would I solve this?

Sketch the triangle, including the third side of unknown length. Since the solution is currently unknown, it doesn't matter whether you sketch an acute or obtuse triangle, only the labels of the parts is important.

The triangle should have one side 6.6 m, the next side 9.8 m, the third side label "x". The angle between the 9.8 m side and the side labeled x is 40 degrees. From the vertex between the two sides measuring 6.6 m and 9.8 m draw an altitude to the side labeled x. There is now a 40 - 50 - 90 right triangle with hypotenuse 9.8 m. The height can be found by using trigonometry: sin 40 = h/9.8 or cos 50 = h/9.8. The side's length, rounded, is 6.3 m.

There are potentially three different lengths for the third side. One length if the original triangle is a right triangle, another if it is acute, the third if it is obtuse. The original triangle is not a right triangle - if it were the altitude would either the third side would have to be the longest side (the two given sides are the legs) or the third side would have to be 6.6 m, which it is not. Therefore, there are two possible lengths for the third side.

The possible side lengths can be found using the law of sines, if that is needed. One sine value relates to two angles, one acute and one obtuse. The two possible angles are approximately 67.4 and 32.6 degrees, and the corresponding side lengths are approximately 9.5 m and 5.5 m

OK-

Let's put the triangle on a cartesian plane. Place the 40 degree angle at the vertex, with the 9.8 m length running along the x-axis. The unknown side lies along a line projecting from the origin -

y = (tan (40) ) x

The 6.6 meter side can be imagined as swinging on a pivot at x=9.8. One can imagine now that part (b) of this problem may be easier to answer than part (a). The 'swinging arm' will intersect the line in two spots. Thus there are two solutions to the problem however we must wait to see if they are both valid solutions...


In order to answer the first part of the problem, we must find the intersection of the circle - defined by the 6.6 m pivoting line segment - with the line. The standard equation for a circle is

x^2 +y^2 = r^2

Here, r = 6.6 m

And, we must translate the circle along the x-axis to x= 9.8... We are left with:

(x-9.8)^2 +y^2 = (6.6)^2

substituting in y= x tan(40)...


(x-9.8)^2 +x^2 (tan(40))^2 = (6.6)^2

expanding and simplifying...

((tan40))%