4.8 - Obtuse angles, elevation and depression

4.8 - Obtuse angles, elevation and depression

You will extend trigonometry to obtuse angles and use angles of elevation and depression to calculate heights and distances. Clear angle geometry matters as much as calculator work. Use degree mode and keep full calculator precision until the final answer.

Trigonometry of obtuse angles (Higher)

An observer's eye is 1.6 m1.6\,\mathrm{m} above level ground, 25 m25\,\mathrm{m} from a tower. The top is seen at 32∘32^\circ above the horizontal. Is 25tan⁡32∘25\tan32^\circ the tower's whole height? We will settle this after extending our angle toolkit.

You already know

For an acute angle in a right triangle, sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Inverse functions recover angles from ratios.

An obtuse angle lies between 90∘90^\circ and 180∘180^\circ. It cannot be a non-right angle inside a right triangle. Instead, extend sine and cosine using a point on a circle of radius 11: a ray turned anticlockwise through θ\theta from the positive horizontal direction meets the circle at (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta).

[DIAGRAM: asset_slug: m071_obtuse_reflection; caption: Acute and obtuse angles reflected in the vertical axis; description: On a unit circle the points at 60 and 120 degrees have equal positive heights and opposite horizontal coordinates. The point at 120 degrees has negative cosine and positive sine.; file: m071_obtuse_reflection.png; recommended_method: matplotlib]
Acute and obtuse angles reflected in the vertical axis

For an obtuse angle, the horizontal coordinate is negative and the vertical coordinate positive. Reflection keeps the height but reverses the horizontal direction. A useful surprise follows: two different angles can have the same sine, such as sin⁡30∘=sin⁡150∘=1/2\sin30^\circ=\sin150^\circ=1/2.

Obtuse-angle relations

For obtuse θ\theta, let its acute reference angle be α=180∘−θ\alpha=180^\circ-\theta. Then

sin⁡θ=sin⁡α,cos⁡θ=−cos⁡α,tan⁡θ=−tan⁡α\sin\theta=\sin\alpha,\qquad \cos\theta=-\cos\alpha,\qquad \tan\theta=-\tan\alpha

Tangent is sine divided by cosine. These relations are absent from the Higher sheet. Not in the formulae sheet: learn it.

Sine is positive for obtuse angles, while cosine and tangent are negative. At 90∘90^\circ, cosine is zero and tangent is undefined; 90∘90^\circ itself is not obtuse.

Worked example 1: exact obtuse values

Question. Find exact values of sin⁡150∘\sin150^\circ, cos⁡150∘\cos150^\circ and tan⁡150∘\tan150^\circ.

Plan. Use the acute reference angle and the signs in the obtuse quadrant.

α=180∘−150∘=30∘reference anglesin⁡150∘=sin⁡30∘=12same vertical coordinatecos⁡150∘=−cos⁡30∘=−32reversed horizontal coordinatetan⁡150∘=−tan⁡30∘=−13sine divided by cosine\begin{aligned} \alpha&=180^\circ-150^\circ=30^\circ &&\text{reference angle}\\ \sin150^\circ&=\sin30^\circ=\frac12 &&\text{same vertical coordinate}\\ \cos150^\circ&=-\cos30^\circ=-\frac{\sqrt3}{2} &&\text{reversed horizontal coordinate}\\ \tan150^\circ&=-\tan30^\circ=-\frac1{\sqrt3} &&\text{sine divided by cosine} \end{aligned}

Why does sine keep its sign? Both circle points lie above the horizontal axis.

Answer. 1/21/2, −3/2-\sqrt3/2 and −1/3-1/\sqrt3, respectively.

Check. Dividing 1/21/2 by −3/2-\sqrt3/2 gives −1/3-1/\sqrt3, as required.

Change only the input angle from 150∘150^\circ to 120∘120^\circ. The sign pattern stays unchanged.

Worked example 2: another obtuse angle

Question. Find exact values of sin⁡120∘\sin120^\circ, cos⁡120∘\cos120^\circ and tan⁡120∘\tan120^\circ.

Plan. Use the new acute reference angle, keeping the obtuse signs.

α=180∘−120∘=60∘reference anglesin⁡120∘=sin⁡60∘=32sine stays positivecos⁡120∘=−cos⁡60∘=−12cosine is negativetan⁡120∘=−tan⁡60∘=−3tangent is negative\begin{aligned} \alpha&=180^\circ-120^\circ=60^\circ &&\text{reference angle}\\ \sin120^\circ&=\sin60^\circ=\frac{\sqrt3}{2} &&\text{sine stays positive}\\ \cos120^\circ&=-\cos60^\circ=-\frac12 &&\text{cosine is negative}\\ \tan120^\circ&=-\tan60^\circ=-\sqrt3 &&\text{tangent is negative} \end{aligned}

Why is the reference angle 60∘60^\circ? It completes a straight angle with 120∘120^\circ, rather than a right angle.

Answer. 3/2\sqrt3/2, −1/2-1/2 and −3-\sqrt3, respectively.

Check. (3/2)/(−1/2)=−3(\sqrt3/2)/(-1/2)=-\sqrt3.

For an input strictly between 00 and 11, inverse sine gives an acute principal angle. An input of 11 gives 90∘90^\circ. If the question asks for an obtuse angle, take the acute angle's supplement. Inverse cosine gives an obtuse angle for an input strictly between −1-1 and 00; an input of −1-1 gives 180∘180^\circ. Inverse tangent of a negative input gives a negative principal angle, so add 180∘180^\circ for the obtuse solution.

Worked example 3: select the obtuse sine solution

Question. Find the obtuse angle θ\theta with sin⁡θ=0.6\sin\theta=0.6, to 11 decimal place.

Plan. Find the acute inverse-sine value, then take its supplement.

α=sin⁡−1(0.6)=36.869897…∘principal acute angleθ=180∘−αchoose the obtuse solution=143.130102…∘\begin{aligned} \alpha&=\sin^{-1}(0.6)=36.869897\ldots^\circ &&\text{principal acute angle}\\ \theta&=180^\circ-\alpha &&\text{choose the obtuse solution}\\ &=143.130102\ldots^\circ \end{aligned}

Why subtract from 180∘180^\circ? Acute and obtuse supplementary angles have equal sine.

Answer. 143.1∘143.1^\circ.

Check. The answer is obtuse, and the sine of the unrounded angle is 0.60.6.

Change only the given sine value to 0.80.8. Keep the requirement that the angle is obtuse.

Worked example 4: a changed sine value

Question. Find obtuse θ\theta with sin⁡θ=0.8\sin\theta=0.8, to 11 decimal place.

Plan. Take the supplement of the principal inverse-sine angle.

α=sin⁡−1(0.8)=53.130102…∘acute candidateθ=180∘−53.130102…∘obtuse candidate=126.869897…∘\begin{aligned} \alpha&=\sin^{-1}(0.8)=53.130102\ldots^\circ &&\text{acute candidate}\\ \theta&=180^\circ-53.130102\ldots^\circ &&\text{obtuse candidate}\\ &=126.869897\ldots^\circ \end{aligned}

Why does the obtuse angle decrease? Its acute supplement increases while their total stays 180∘180^\circ.

Answer. 126.9∘126.9^\circ.

Check. The unrounded angle has sine 0.80.8 and lies between 90∘90^\circ and 180∘180^\circ.

For the other inverse functions, the same sign picture controls the answer. For example, cos⁡−1(−0.6)=126.869897…∘\cos^{-1}(-0.6)=126.869897\ldots^\circ is already obtuse. If tan⁡ϕ=−0.8\tan\phi=-0.8, the obtuse value is 180∘+tan⁡−1(−0.8)=141.340191…∘180^\circ+\tan^{-1}(-0.8)=141.340191\ldots^\circ, or 141.3∘141.3^\circ to 11 decimal place. Its tangent is negative, as the picture predicts.

Watch out

The line θ=sin⁡−1(0.6)=36.9∘\theta=\sin^{-1}(0.6)=36.9^\circ ignores the word “obtuse”. The inverse gives the reference angle; use 180∘−sin⁡−1(0.6)180^\circ-\sin^{-1}(0.6) for the requested angle.

Check both the sign and the stated angle range, rather than accepting the first display value.

Angles of elevation and height (Higher)

An angle of elevation is measured upwards from the observer's horizontal to the line of sight. It is not measured from a vertical wall. On level ground, a vertical object gives a right triangle, but its vertical side is the rise above the observer's eye.

[DIAGRAM: asset_slug: m071_elevation_depression; caption: Elevation and depression angles from parallel horizontals with eye height e and total height h; description: The line OT joins eye O to tower top T. Elevation at O equals depression at T because the horizontals are parallel. Tower height h includes eye height e while the vertical rise in the sight triangle is h minus e. Diagram is schematic.; file: m071_elevation_depression.png; recommended_method: matplotlib]
Elevation and depression angles from parallel horizontals with eye height e and total height h

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