Triganometry question.?

Question

Please help:

look at this triangle http://i104.photobucket.com/albums/m174/tomw91/triangle.jpg

a.) Calculate the distance x.

b.) At its shortest distance from the top angle, a boat is at point z on the line x. Find the distance from the top angle to this point (to 3 significant figures.)

Please show me how to do this so i can apply the method to my other questions.

Answer 1

[Edit: Timmy, above, had the right idea, but he got a sign wrong. Instead of writing 36+25, he inadvertently wrote 36-25. That messed up his result, but he knew what he was doing. End edit.]

Label some points on your diagram. Put A at your lower left, B at the top (at the 150-degree angle), C at the lower right, and let your "point z" be labeled D. Then AB = 5, BC = 6, AC = x, and we can say BD = z.

The segment BD, by the way, is perpendicular to AC.

1. Calculate the distance x. Use the Law of Cosines:

AC^2 = AB^2 + BC^2 - 2 AB BC cos ABC

x^2 = 5^2 + 6^2 - 2*5*6 cos 150
x^2 = 25 + 36 - 60 (-sqrt 3 / 2) = 61 + 30 sqrt 3 = 112.9615
x = 10.628 m

2.Find the distance z.

Notice that ADB and CDB are both right triangles. For clarity, let the unknown distance AD be labeled y. Then, CD = x-y (and, of course, we already know the value of x.) (And as a reminder, BD = z.)

In triangle ADB, we have

y^2 + z^2 = 5^2
z^2 = 25 - y^2 (Equation 1)

and in triangle BDC, we have

(x-y)^2 + z^2 = 6^2
z^2 = 36 - (x^2 - 2xy + y^2)
z^2 = 36 - x^2 + 2xy - y^2 (Equation 2)

We want to solve Equations 1 and 2 for z. First, set them equal to each other:

25 - y^2 = 36 - x^2 + 2xy - y^2
x^2 - 11 = 2xy
y = (x^2 - 11) / 2x = 4.79668 (when I plugged in the value for x)

Now substitute this y-value into Equation 1:

z^2 = 25 - y^2 = 1.99183

z = 1.411 (Answer)

Check:

y^2 + z^2 = 4.79668^2 + 1.99183 = 25 (AB^2 -- OK)
(x-y)^2 + z^2 = 5.83165^2 + 1.99183 = 36 (BC^2 -- OK)

The answer is correct, and I think you can figure out the method.

Answer 2

a) Since this is a non-right-angle triangle, we need the sine or cosine law. Since you know one angle and two sides, use this form of the cosine law:

a^2 = b^2 + c^2 - 2 b c cos(A)

where
a=x
b=6km
c=5km
A=150°

x^2 = 6^2 + 5^2 - 2(6)(5)cos(150°)
x^2 = 36 - 25 - (60)(-0.866)
x^2 = 62.96
x = 7.93km

b) At its shortest distance from the top angle on line x, the boat would be at a point where a vertical line from the top angle hits line x. That will form a right-angle with x, and we can split the triangle into two right-angle triangles. Let's call the distance from the top angle to this point m and the distance along x from the left-hand side where it intersects n. Then we can write two right-angle triangle equations with two variables each and solve.

n^2 + m^2 = 5^2
(7.93-n)^2 + m^2 = 6^2

From the first, m^2 = 25-n^2
Sub that in the second, and you get:

(7.93-n)^2 + (25-n^2) = 36
(62.885 - 15.86n + n^2) + 25 - n^2 = 36
-15.86n = 36 - 62.885 - 25
n = 3.459

which makes m, the distance of the ship from the angle = 3.610km

Answer 3

I think x comes out to be

√[a² + b² + ab sin Θ]

They're basically asking you for the height of the segment, perpendicular to x, that passes through z. Divide the triangle into two right triangles, and use trig functions on both (sin) to find that segment's length. You're given the two hypoteneuses, and the sum of the two angles (150º) so it shouldn't be too hard.

Answer 4

The horizontal asymptotes are whilst the real of the fraction equals 0 and that's at 3. The vertical asymptotes are whilst the backside of the fraction equals 0 and that's at +3 and -3. wish this facilitates