I’m prepping for a geometry test and my brain is doing little cartwheels over congruent triangles.
Here’s the setup from my worksheet: Triangle PQR has PQ = 6 cm, QR = 8 cm, and angle P = 40°. Triangle XYZ has XY = 6 cm, YZ = 8 cm, and angle X = 40°. Both 40° angles touch the 6 cm side, but the 8 cm side is kind of opposite/adjacent in different ways depending on the drawing. I keep second-guessing what’s “included.”
My attempt: I wrote SAS at first (two sides match and an angle matches!), then realized the angle might not be the included one, so maybe that’s wrong. I tried rotating/flipping one triangle onto the other, but when I sketch, I can actually make two different triangles that still fit 6, 8, and 40°-one sort of bulges left, the other right. So now I’m thinking this is that notorious “SSA” trap?
Questions:
– With just 6, 8, and a 40° angle not necessarily between them, are these triangles definitely congruent, or could there be more than one possibility?
– If they are congruent, how do I correctly match the vertex order (like ΔPQR ≅ ΔXYZ or ΔXZY)? I always mess up which vertices correspond.
– If they’re not, what specific extra fact (which angle or which side placement) would lock in congruence?
Tiny side-quest: for right triangles, does the hypotenuse-leg thing avoid this included-angle confusion, or am I mixing rules again?
Would love a nudge in the right direction.
Ready to make maths more enjoyable, accessible, and fun? Join a friendly community where you can explore puzzles, ask questions, track your progress, and learn at your own pace.
By becoming a member, you unlock:
3 Responses
SSA with 6, 8, and a 40° not included is ambiguous-two triangles are possible unless the 8 cm side is opposite the 40°, in which case it’s unique; to force congruence use SAS (treating the 40° next to the 6 cm as the included angle) or ASA/AAS, match vertices by the equal 40° angles and the 8 cm sides, and note that HL for right triangles avoids this issue (see https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-law-of-sines/a/ambiguous-case-of-the-law-of-sines).
Hope this helps!
SSA is the carnival-mirror case: it can give 0, 1, or 2 triangles-BUT in your labeled setup the 8 cm side is opposite the 40° and 8 > 6, so there’s exactly one triangle and ΔPQR ≅ ΔXYZ with P↔X (40°), Q↔Y (side 6), R↔Z (side 8). To lock congruence more generally, make the 40° the included angle with 6 and 8 (SAS) or add another angle (ASA/AAS), and for right triangles HL works too (basically SAS for right triangles-tiny cheat); nice walkthrough: https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-law-of-sines/a/law-of-sines-ambiguous-case – Hope this helps!
Short answer: with your numbers there’s only one triangle. You’ve got an angle of 40° with the side next to it equal to 6, and the side opposite it equal to 8. In the SSA “ambiguous case,” two different triangles only happen when the opposite side is shorter than the adjacent side but still longer than the altitude h = (adjacent)·sin(angle). Here, h = 6·sin 40° ≈ 3.86, and the opposite side is 8, which is not just longer than h but also longer than 6. That puts you firmly in the “one triangle” bucket. The two sketches that “bulge left/right” are just mirror images-still congruent.
To match vertices: angle P = angle X (both 40°), and the sides opposite those angles match too (QR = 8 corresponds to YZ = 8). The 6 cm side touching the 40° is PQ in the first triangle and XY in the second, so Q ↔ Y. That forces R ↔ Z. So ΔPQR ≅ ΔXYZ. If you ever do want to “lock in” congruence when SSA might be ambiguous, you need one of these: make the 40° the included angle between the 6 and 8 (that’s SAS), or give a second angle (ASA/AAS), or tell me explicitly that the side opposite the given acute angle is the longer of the two given sides (which kills the ambiguity). Tiny side-quest: yes, the right-triangle hypotenuse–leg rule is legit and neatly avoids the “included angle” fuss-it’s the one place where SSA behaves.
When I first learned this, I kept drawing the “left bump” and “right bump” pictures and swore they were different triangles. My teacher had me do one quick check: compute h = (adjacent)·sin(given angle) and compare. If opposite < h: no triangle. If opposite = h: exactly one (right) triangle. If h < opposite < adjacent: two triangles. If opposite ≥ adjacent: exactly one. Once I started that 5-second test, the SSA drama evaporated.