Obtuse Angles and Exact Values
Trigonometric ratios also work for obtuse angles: \(\sin(180^\circ-\theta)=\sin\theta\), while cosine and tangent change sign, so \(\cos 140^\circ=-\cos 40^\circ\). The angles \(30^\circ\), \(45^\circ\) and \(60^\circ\) have exact ratios with surds. In NSW Year 10 Advanced maths you use both ideas to find exact values and solve equations (NESA outcome MA5‑TRG‑P‑02).
You will write the ratios of obtuse angles using acute angles, find exact values such as \(\tan 150^\circ\) and \(\cos 135^\circ\), and use complementary angles. You will find exact lengths and areas, solve an equation like \(\sin\theta=0.8\) for both angles between \(0^\circ\) and \(180^\circ\), find the angle a line makes with the \(x\)-axis, and use one ratio of an obtuse angle to find the other two. It is part of Year 10 Advanced Trigonometry.
Theory
For an obtuse angle \(\theta\), use the related acute angle \(180^\circ - \theta\): sine keeps its sign, while cosine and tangent become negative. This Year 10 Advanced maths page also covers the exact values for \(30^\circ\), \(45^\circ\) and \(60^\circ\), complementary angles, solving equations between \(0^\circ\) and \(180^\circ\), and exact answers with surds. It leads straight into Mathematics Advanced in Years 11 and 12.
An obtuse angle is between \(90^\circ\) and \(180^\circ\). Its trigonometric ratios come from the related acute angle \(180^\circ - \theta\):
\(\sin(180^\circ - \theta) = \sin\theta\), \(\cos(180^\circ - \theta) = -\cos\theta\), \(\tan(180^\circ - \theta) = -\tan\theta\).
On a circle of radius \(1\), the point at angle \(\theta\) is \((\cos\theta, \sin\theta)\). An obtuse angle points into the second quadrant, where \(x\) is negative and \(y\) is positive. That is why only sine is positive there. At the boundaries: \(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\), \(\cos 180^\circ = -1\), and \(\tan 90^\circ\) is undefined.
Exact values are surd or fraction answers, such as \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\), found from two special triangles without a calculator. Complementary angles add to \(90^\circ\): \(\sin A = \cos(90^\circ - A)\) and \(\cos A = \sin(90^\circ - A)\), so \(\cos 63^\circ = \sin 27^\circ\).
Exact values:
| \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | |
|---|---|---|---|
| \(\sin\) | \(\dfrac{1}{2}\) | \(\dfrac{1}{\sqrt{2}}\) | \(\dfrac{\sqrt{3}}{2}\) |
| \(\cos\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{\sqrt{2}}\) | \(\dfrac{1}{2}\) |
| \(\tan\) | \(\dfrac{1}{\sqrt{3}}\) | \(1\) | \(\sqrt{3}\) |
How to find a ratio of an obtuse angle
- Find the related acute angle \(180^\circ - \theta\).
- Write the ratio of that acute angle, using the exact-value table if it is \(30^\circ\), \(45^\circ\) or \(60^\circ\).
- Attach the sign: \(+\) for sine, \(-\) for cosine and tangent.
How to solve an equation for \(0^\circ \le \theta \le 180^\circ\)
- Make the ratio the subject, e.g. \(2\cos\theta + 1 = 0\) gives \(\cos\theta = -\dfrac{1}{2}\). For \(\sin^2\theta = k\), take the square root; only the positive root of \(\sin\theta\) is possible in this range.
- Find the related angle from the positive value (calculator or exact values).
- Decide the quadrant: sine gives \(\alpha\) and \(180^\circ - \alpha\); negative cosine or tangent gives \(180^\circ - \alpha\) only.
Exact-answer problems
- Given one ratio, find the others: draw the triangle for the related angle, find the third side with Pythagoras' theorem, then attach the signs (Example 3).
- Exact lengths: use exact values in place of decimals, rationalise denominators, and simplify surds. To get a right angle next to an obtuse angle, extend a side (Example 4).
- Show that: work on one side at a time with exact values until both sides match.
(i) \(\tan 150^\circ\)
(ii) \(\cos 135^\circ\)
(iii) \(\sin 45^\circ\cos 135^\circ - \cos 60^\circ\)
Use the related acute angle \(180^\circ - \theta\). Sine stays positive; cosine and tangent change sign.
(i) Related angle \(30^\circ\):
| \(\tan 150^\circ\) | \(=\) | \(-\tan 30^\circ\) |
| \(\) | \(=\) | \(-\dfrac{1}{\sqrt{3}} = \mathbf{-\dfrac{\sqrt{3}}{3}}\) |
(ii) Related angle \(45^\circ\):
| \(\cos 135^\circ\) | \(=\) | \(-\cos 45^\circ\) |
| \(\) | \(=\) | \(-\dfrac{1}{\sqrt{2}} = \mathbf{-\dfrac{\sqrt{2}}{2}}\) |
(iii) Substitute the exact values:
| \(\) | \(\) | \(\dfrac{1}{\sqrt{2}} \times \left(-\dfrac{1}{\sqrt{2}}\right) - \dfrac{1}{2}\) |
| \(\) | \(=\) | \(-\dfrac{1}{2} - \dfrac{1}{2}\) |
| \(\) | \(=\) | \(\mathbf{-1}\) |
Rationalise the denominator: multiply top and bottom by the surd, \(\dfrac{1}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\).
(i) \(\sin\theta = 0.35\)
(ii) \(\tan\theta = -2.4\)
(iii) \(2\sin\theta = \sqrt{2}\)
(i) Sine is positive for acute and obtuse angles, so there are two answers.
| \(\theta\) | \(=\) | \(\sin^{-1}(0.35) = 20.487\ldots^\circ\) |
| \(\text{or } \theta\) | \(=\) | \(180^\circ - 20.487\ldots^\circ = 159.512\ldots^\circ\) |
| \(\theta\) | \(\approx\) | \(\mathbf{20^\circ}\text{ or }\mathbf{160^\circ}\) |
(ii) Tangent is negative, so \(\theta\) is obtuse. Find the related angle from the positive value, then subtract from \(180^\circ\).
| \(\tan\alpha\) | \(=\) | \(2.4,\quad \alpha = 67.380\ldots^\circ\) |
| \(\theta\) | \(=\) | \(180^\circ - \alpha = 112.619\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{113^\circ}\) |
(iii) \(\sin\theta = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\), an exact value with related angle \(45^\circ\).
| \(\theta\) | \(=\) | \(\mathbf{45^\circ}\text{ or } 180^\circ - 45^\circ = \mathbf{135^\circ}\) |
A calculator gives only one answer for \(\sin^{-1}\). The second one, \(180^\circ - \theta\), is your job.
Draw the triangle for the related acute angle \(\alpha = 180^\circ - \theta\), with \(\tan\alpha = \dfrac{8}{15}\). Ignore the sign while you draw.
| \(\text{hypotenuse}\) | \(=\) | \(\sqrt{8^2 + 15^2} = \sqrt{289} = 17\) |
Now put the signs back. For an obtuse angle, sine is positive and cosine is negative.
| \(\sin\theta\) | \(=\) | \(\mathbf{\dfrac{8}{17}}\) |
| \(\cos\theta\) | \(=\) | \(\mathbf{-\dfrac{15}{17}}\) |
Check: \(\dfrac{\sin\theta}{\cos\theta} = \dfrac{8}{17} \div \left(-\dfrac{15}{17}\right) = -\dfrac{8}{15}\).
\(C\), \(B\) and \(D\) are on a straight line, so \(\angle ABD = 180^\circ - 135^\circ = 45^\circ\). Triangle \(ABD\) is right-angled at \(D\).
| \(AD\) | \(=\) | \(6\sin 45^\circ = 6 \times \dfrac{1}{\sqrt{2}} = \mathbf{3\sqrt{2}}\text{ cm}\) |
| \(BD\) | \(=\) | \(6\cos 45^\circ = \mathbf{3\sqrt{2}}\text{ cm}\) |
Now use Pythagoras' theorem in triangle \(ACD\), with \(CD = 4\sqrt{2} + 3\sqrt{2} = 7\sqrt{2}\).
| \(AC^2\) | \(=\) | \((7\sqrt{2})^2 + (3\sqrt{2})^2\) |
| \(\) | \(=\) | \(98 + 18 = 116\) |
| \(AC\) | \(=\) | \(\sqrt{116} = \mathbf{2\sqrt{29}}\text{ cm}\) |
Simplify the surd: \(\sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29}\).
Common pitfalls
Frequently asked questions
Why is sin of an obtuse angle positive?
On a circle of radius 1, sine is the height of the point at that angle. An obtuse angle points into the second quadrant, which is above the x-axis, so the height is positive. Cosine, the horizontal value, is negative there.
How do you find the exact value of sin 135 degrees?
The related acute angle is \(180^\circ - 135^\circ = 45^\circ\). Sine stays positive for an obtuse angle, so \(\sin 135^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\).
What are the exact values of sin, cos and tan for 30, 45 and 60 degrees?
\(\sin 30^\circ = \dfrac{1}{2}\), \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\), \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\). Cosine runs the other way. \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\), \(\tan 45^\circ = 1\), \(\tan 60^\circ = \sqrt{3}\). The table in the Formulas tab lists them all.
Why does sin x = 0.5 have two answers?
Between \(0^\circ\) and \(180^\circ\), sine is positive for both acute and obtuse angles, and \(\sin(180^\circ - x) = \sin x\). So \(\sin x = 0.5\) gives \(x = 30^\circ\) or \(x = 150^\circ\).
What does sin A = cos(90 - A) mean?
In a right-angled triangle the two acute angles add to \(90^\circ\), and the side opposite one is adjacent to the other. So the sine of an angle equals the cosine of its complement, for example \(\sin 20^\circ = \cos 70^\circ\).