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RightTriangle Word Problems Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it. Once you've got a helpful diagram, the math is usually pretty straightforward.
For the base, I'll use the cosine ratio: cos(60°) = b/6
The ladder reaches
about 5.20 meters up the wall, Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round in the exercise above, my value for the height should have been the value with the radical.
They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios. sin(α) = 4/5 The angle the ladder makes with the ground is about 53°.
I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio: Copyright © Elizabeth Stapel 20102011 All Rights Reserved tan(17/60°) =
b/92919800 This is just half the width; carrying the calculations in my calculator (to minimize roundoff error), I get a value of 919002.8129. This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise. The diameter is about 919,003 miles.
The bearings tell me the angles from "due north", in a clockwise direction. Since 130 – 40 = 90, these two bearings will give me a right triangle. From the times and rates, I can find the distances: 1.3 × 110 = 143 Now that I have the lengths of the two
legs, I can set up a triangle: I can find the distance by using the Pythagorean Theorem: 143^{2} + 165^{2}
= c^{2}
The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure: 165/143 = tan(θ)
But this is not the "bearing", since the bearing is the angle with respect to "due north". I need to add in the original fortydegree angle to get my answer: The plane is about 218 miles away, at a bearing of about 89°. Another major class of righttriangle word problems you will likely encounter is angles of elevation and depression....



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