For an obtuse angle θ between 90° and 180°, sin θ is positive and equals sin(180° − θ), while cos θ and tan θ are negative and equal −cos(180° − θ) and −tan(180° − θ). Work out the ratio for the reference angle 180° − θ first, then attach the correct sign.
Up to this point, trigonometric ratios have always come from a right-angled triangle, where every angle is between 0° and 90°. An obtuse angle, between 90° and 180°, cannot sit inside a right-angled triangle at all, so the usual SOH-CAH-TOA picture breaks down. Students often assume the ratios simply do not exist for obtuse angles, or they plug the angle into a calculator, get a negative number for cosine, and panic without understanding why the sign flipped.
The syllabus needs obtuse-angle ratios for two reasons: the sine rule and cosine rule both throw up obtuse angles in non-right-angled triangles, and questions on y = sin x style graphs test whether you understand the pattern across the full 0° to 180° range.
Step 1
Find the reference angle: 180° − θ
Step 2
Work out sin, cos, tan of the reference angle
Step 3
sin θ = sin(180° − θ), stays positive
Step 4
cos θ = −cos(180° − θ), tan θ = −tan(180° − θ)
Find sin 130°, cos 130° and tan 130°, correct to 3 significant figures.
Solution
180° − 130° = 50°.sin 50° = 0.766, cos 50° = 0.643, tan 50° = 1.19 (3 s.f.).sin 130° = sin 50° = 0.766 (3 s.f.).cos 130° = −cos 50° = −0.643 and tan 130° = −tan 50° = −1.19 (3 s.f.).Quick check: type sin 130°, cos 130° and tan 130° directly into your calculator in degree mode. The values should match exactly. This confirms the identity without you needing to memorise anything beyond "sine stays positive, cosine and tangent flip".
Given that sin θ = 0.6 and 0° ≤ θ ≤ 180°, find both possible values of θ.
Solution
θ = sin⁻¹(0.6) = 36.9° (3 s.f.).sin θ = sin(180° − θ) to find it: θ = 180° − 36.9° = 143.1° (1 d.p.).θ = 36.9° or θ = 143.1°.The step students get wrong
When a question gives sin θ = k with θ allowed up to 180°, the calculator's sin⁻¹ button only ever returns the acute answer. Students who stop after step 1 lose the obtuse solution entirely, which is usually worth marks on its own. Whenever the range includes angles beyond 90° and you are solving using sine, always check whether a second, obtuse answer is required using 180° − θ.
Obtuse angles show up constantly once you move from right-angled triangles to any triangle. When using the sine rule to find an angle, the same ambiguous-case issue from worked example 2 applies: sine gives you an acute answer first, and you must check the triangle's shape (using the given side lengths, or knowing which angle is largest) to decide whether the obtuse alternative is the correct one. The cosine rule avoids this problem entirely, because cosine only returns one angle between 0° and 180° for a valid triangle: if the calculation gives a negative cosine, the angle you are finding is obtuse, and the calculator handles the conversion automatically.
Obtuse angles also appear directly inside the area of a triangle formula, Area = ½ab sin C, which works correctly for an obtuse angle C without any adjustment, precisely because sin C stays positive throughout 0° to 180°.
Exam tip: if a cosine rule calculation for an angle produces a negative value inside cos⁻¹, do not treat this as an error. A negative cosine simply means the angle is obtuse, and your calculator will return the correct obtuse angle directly, no extra step needed.
The identities in this guide are exactly why the graph of y = sin x is symmetrical about x = 90° between 0° and 180°, rising to a peak of 1 at x = 90° and falling back down, always positive. The graph of y = cos x, by contrast, starts at 1 when x = 0°, crosses zero at x = 90°, and continues into negative values all the way to x = 180°. Recognising these two shapes makes it easy to sanity-check any obtuse-angle ratio you calculate: if your value for sine of an obtuse angle came out negative, or your value for cosine came out positive, you have made a sign error somewhere.
Why is sine positive but cosine negative for obtuse angles?
This comes from the unit circle definition of trigonometric ratios: for an angle between 90° and 180°, the point on the unit circle has a positive y-coordinate (giving positive sine) but a negative x-coordinate (giving negative cosine). Tangent, being sine divided by cosine, then works out negative too, since a positive divided by a negative is negative.
Do I need to know the unit circle for O-Level E-Maths?
No. The syllabus only requires you to apply the identities sin θ = sin(180° − θ), cos θ = −cos(180° − θ) and tan θ = −tan(180° − θ), and to use your calculator correctly for obtuse angles. Understanding where the identities come from helps you remember them, but you will not be asked to derive them from the unit circle.
Can I just type the obtuse angle straight into my calculator?
Yes, for finding a ratio from a known angle your calculator handles obtuse angles correctly in degree mode. The identities matter most for solving equations, where the calculator's inverse trig functions only return one answer and you must work out whether a second, obtuse solution is also needed.
Is 90° itself acute or obtuse?
90° is neither acute nor obtuse, it is a right angle. At exactly 90°, sin 90° = 1 and cos 90° = 0, which is the boundary point where the sign of cosine switches from positive to negative as the angle increases past it. Because cos 90° = 0, tan 90° is undefined, so your calculator returns an error for it: that is the one angle in this range where the tangent identity has nothing to give you.
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