Finding a Missing AngleDate: 01/23/2002 at 21:11:07 From: Mel Subject: Area of a triangle I am home schooling and am stuck on an assignment about geometry, and trigonometry. Here's one question: I have a triangle with sides: AB = 3.6 cm BC = 6.5 cm CA = 5.5 cm and angle CAB is 90 degrees. The question is: Using trigonometry, calculate the measure of angles ABC and ACB. How can I find the missing angle? If I had two angles I could subtract them from 180 degrees, but I only have one. And I don't know how to find the area of a triangle. In my assignment a triangle measures: CA = 10.8cm AB = 7.2cm BC = 13cm and it says find the area. But I don't know how. Can you help me? Thanks in advance, Mel Date: 01/24/2002 at 20:11:06 From: Doctor Jeremiah Subject: Re: Area of a triangle Hi Mel, Trigonometry is the way you find angles when you know only one angle. There are only three important things you need to know to do trigonometry. First, the triangle has to have a 90-degree angle. Second, you need to know the definitions for sine, cosine, and tangent. And third, in order to make those definitions you need to define a couple of other things: the hypotenuse, the adjacent side, and the opposite side. The triangle MUST have a 90-degree angle. And what good luck, yours does! C + /| / | / | / | / | 6.5 5.5 / | / | / | +---3.6---+ B A The hypotenuse is the side that does not touch the 90-degree angle. It is also the longest side. In your triangle it is side BC and has a length of 6.5 The adjacent side and the opposite side move around depending on what angle you want to calculate. Say you want to find the value of angle CBA. The adjacent side is the side that touches angle CBA (and isn't the hypotenuse). In this case it is side AB because side AB touches angle CBA and side AC does not. The opposite side is the side that does not touch angle CBA (and also isn't the hypotenuse). In this case it is side AC because side AC does not touch angle CBA and side AB does. And if you are trying to find angle BCA then BC is the hypotenuse, AB is the opposite side, and AC is the adjacent side. So what are these definitions good for? Well now we can define the sine, cosine, and tangent: sine: sin(angle) = opposite/hypotenuse cosine: cos(angle) = adjacent/hypotenuse tangent: tan(angle) = opposite/adjacent Lets say you want to find the size of angle CBA. The hypotenuse is BC, the adjacent side is AB, and the opposite side is AC. We choose an equation from among sine, cosine, and tangent depending on what information we know. If we know only two sides, we will pick the equation that uses those two sides. But we know all three sides so it doesn't matter. So we will pick one: sin(angle) = opposite/hypotenuse sin(CBA) = AC/BC sin(CBA) = 5.5/6.5 sin(CBA) = 0.846 Now how do we change this into something that equals angle CBA? Right now it's the sine of CBA and that is not useful. Well, you know how the square and square root cancel each other out: sqrt(x squared) = x. It turns out that there are also functions that undo sine, cosine, and tangent. They are called the arcsine, arccosine, and arctangent: arcsine: arcsin( sin(angle) ) = angle arccosine: arccos( cos(angle) ) = angle arctangent: arctan( tan(angle) ) = angle So when we have an equation that looks like: sin(CBA) = 0.846 We can do this: sin(CBA) = 0.846 arcsin( sin(CBA) ) = arcsin( 0.846 ) CBA = arcsin( 0.846 ) Now, to do the arcsin you will need a calculator or a table or arcsines or a slide rule. On a calculator the arcsine is sometimes called the inverse sine, or sometimes has the symbol of sin with a -1 exponent, and sometimes it's unlabeled but is on the secondary function ogf the sine button. For your second question, about triangle area: When you don't know for sure that the triangle has a 90-degree angle, you have to be sneaky. If you know for sure that it has a 90-degree angle, then it's just half a rectangle (cut diagonally). But without a 90-degree angle we need to use Heron's formula. Heron also proves his famous formula: If A is the area of a triangle with sides a, b, and c, and s = (a+b+c)/2, then A = sqrt[s(s-a)(s-b)(s-c)]. Let's change your triangle to use sides named a, b and c: a = 7.2cm b = 10.8cm c = 13cm Now: s = (a+b+c)/2 s = (7.2+10.8+13)/2 s = 15.5 And: A = sqrt[ s(s-a)(s-b)s-c) ] A = sqrt[ 15.5 (15.5 - 7.2) (15.5 - 10.8) (15.5 - 13) ] A = sqrt[ 15.5 (8.3) (4.7) (2.5) ] A = 38.87978 Let me know if you still have questions. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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