# Two trig questions (Triangles and Trapezoid)

1. Feb 9, 2009

### rought

1. The problem statement, all variables and given/known data

An isosceles trapezoid has a height of 4 cm and bases 3 cm and 7 cm long. How long are its diagonals?

From lighthouses P and Q, 16km apart, a disabled ship S is sighted. If angle SPQ = 44 and angle SQP = 66, find the distance from S to the nearer lighthouse?

I know how to do all of the calculations but im just not sure where to start on these =/ ?

2. Feb 9, 2009

### AssyriaQ

Well, you need to show some work or at least show your thoughts on how to proceed. What have you done so far?

Hint/question: What is nice about the trapezoid being isosceles?

3. Feb 9, 2009

### Staff: Mentor

For #1, draw a picture of the trapezoid on a coordinate axis system. I would center the long base on the x axis. Find the coordinates of the four vertices and calculate the distance from the opposite vertices. They should be equal.

For #2, draw a picture of the triangle formed by the two lighthouses and the ship, and label the vertices as P, Q, and S (the ship). Label the side you know and the angles you know. From the given information, use what you know about trig to find the missing side. You'll probably need to use the Law of Sines.

4. Feb 9, 2009

### rought

Alright well I am good with the second question on the ship i am pretty sure i got it...

For the trapezoid here's what I drew

I'm not sure where to go from here

5. Feb 9, 2009

### Staff: Mentor

Look a bit closer at what I said earlier:

6. Feb 9, 2009

### rought

Alright I graphed it out, with (0,0) and (0,7) being the two bottom vertices, and (2,4) and (5,4) for the others... would the distance be 4? Between one of the long vertices and one of the short?

7. Feb 9, 2009

### AssyriaQ

Just use the Pythagorean theorem to calculate the diagonal (which would be the hypotenuse). You already know the height, and you can easily find out the other cathetus (you more or less already have it since you wrote out coordinates).

8. Feb 9, 2009

### rought

ahh ok, √5² + 4² = √41 ≈ 6.403 that seem right?

9. Feb 9, 2009

### AssyriaQ

Correct.

10. Feb 10, 2009

### HallsofIvy

Why 5? The right triangle with the lower right corner as one vertex and the diagonal as hypotenuse has height 4 cm, the height of the parallelogram, and base 7- 3= 4 cm.

11. Feb 10, 2009

### AssyriaQ

No, the base is 7-2=5 cm.

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