I’m stuck on how to get the area of an obtuse triangle when the perpendicular height doesn’t land on the base segment. I’ve got a triangle where the base is 14 cm, the side next to it is 9 cm, and the angle between them is 120°. The little right-angle drop from the opposite vertex would hit the extension of the base, not the base itself, which is where my brain short-circuits.
Confession: I’ve struggled with triangle areas since forever – I once turned in a whole worksheet without the “divide by 2” part. My teacher used to say the height is “shy” and sometimes hides outside the triangle. Cute, but I still can’t turn that into numbers.
Here’s my (completely wrong) attempt: I panicked and did (14 + 9 + 120)/2 = 71.5 and called that the area. I know… that mixes units and makes no sense, but it looked satisfyingly math-y for five seconds.
Can someone explain, step by step, how to get the actual area in this setup? Do I extend the base and just use the length of that outside perpendicular as the height? With base = 14 cm, adjacent side = 9 cm, and included angle = 120°, what’s the right way to compute the height and then the area?
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2 Responses
Yep – you just extend the base and use the perpendicular distance to that line as the height. The half-base-times-height formula doesn’t care if the foot lands outside the segment. In your setup, take the side of 9 cm making 120° with the base. The height relative to the base is h = 9·sin(120°). Since sin(120°) = sin(60°) = √3/2, you get h = 9·√3/2 ≈ 7.79 cm. Then area = 1/2 · 14 · h = 1/2 · 14 · (9√3/2) = 31.5√3 ≈ 54.6 cm². Shortcut: when you know two sides and the included angle, you can jump straight to A = (1/2)ab·sin(C). Here that’s (1/2)·14·9·sin(120°), same result.
I used to get tripped up by the “height outside” thing too. My fix was to remember two quick checks: cos(120°) is negative (so the foot lands outside), but sin(120°) is positive (so the height is just the “up” component). In other words, treat an obtuse angle like its supplement for sine. Once I started doing “half × side × side × sine of the included angle,” I stopped second-guessing and my triangles stopped looking like mystery blobs. And yes, I once forgot the divide-by-2 on a whole quiz – looked satisfyingly math-y for five seconds there too.
Totally fine that the “shy” height lands outside-just take the perpendicular component of the 9 cm side: h = 9 sin(120°), so area = (1/2) * 14 * h = (1/2) * 14 * 9 * sin(120°) = 63 * (√3/2) ≈ 54.6 cm² (nice walkthrough: https://www.mathsisfun.com/algebra/trig-area-triangle-with-sine.html).
Does thinking of it as “drop the altitude to the extended base, so h = opposite side × sin(included angle)” make it click, or want a quick sketch/derivation?