C6.2; E6.2 - Finding angles in right triangles

C6.2; E6.2 - Finding angles in right triangles

Side lengths can reveal an angle when you reverse a trigonometric ratio. You will choose an inverse sine, cosine or tangent calculation, find the acute angle in degrees, and keep enough precision to round the final angle correctly.

Choosing sides for an inverse ratio

A right triangle has opposite side 6 cm6\,\mathrm{cm} and adjacent side 8 cm8\,\mathrm{cm}. If you double both lengths, will the angle change? The ratio stays 6/8=3/46/8=3/4, so the acute angle stays the same. To discover that angle, reverse the tangent calculation.

If tan⁡θ=0.75\tan\theta=0.75, you need the acute angle whose tangent is 0.750.75. On a scientific calculator, the inverse tangent key is usually written tan⁡−1\tan^{-1}, often accessed using SHIFT or a similar key. In degree mode it gives 36.8698…∘36.8698\ldots^\circ.

Inverse trigonometric functions

The inverse sine, inverse cosine and inverse tangent functions recover an angle from its ratio. For an acute angle in a right triangle,

θ=sin⁡−1(oh)θ=cos⁡−1(ah)θ=tan⁡−1(oa)\begin{aligned} \theta&=\sin^{-1}\left(\frac{o}{h}\right)\\ \theta&=\cos^{-1}\left(\frac{a}{h}\right)\\ \theta&=\tan^{-1}\left(\frac{o}{a}\right) \end{aligned}

Here o,a,ho,a,h mean opposite, adjacent and hypotenuse relative to θ\theta. Use the one formula containing the two known sides.

Sine and cosine ratios lie strictly between 00 and 11 for an acute angle. Tangent is positive and can exceed 11: a steeper triangle can have opposite side longer than adjacent. These conditions help detect a reversed fraction.

[DIAGRAM: asset_name: Choosing known sides for an inverse ratio; asset_slug: trig_inverse_known_pairs; caption: Choose the inverse ratio from the two known sides.; description: The same six-eight-ten right triangle is shown twice with the same target angle theta. The left copy supplies opposite six and adjacent eight; the right supplies opposite six and hypotenuse ten. This changes the available ratio while preserving the angle.]
Choose the inverse ratio from the two known sides.

Worked example 1: two shorter sides

Question. A right triangle has opposite side 6 cm6\,\mathrm{cm} and adjacent side 8 cm8\,\mathrm{cm}, relative to θ\theta. Calculate θ\theta, correct to 11 decimal place.

Plan. Use tangent for opposite and adjacent, then its inverse in degree mode.

tan⁡θ=68the two known shorter sidesθ=tan⁡−1(68)recover the acute angle=36.8698…∘calculator in degree mode\begin{aligned} \tan\theta&=\frac68 && \text{the two known shorter sides}\\ \theta&=\tan^{-1}\left(\frac68\right) && \text{recover the acute angle}\\ &=36.8698\ldots^\circ && \text{calculator in degree mode} \end{aligned}

Answer. θ=36.9∘\theta=36.9^\circ.

Check. Pythagoras gives hypotenuse 62+82=10\sqrt{6^2+8^2}=10. The alternative calculation sin⁡−1(6/10)\sin^{-1}(6/10) also gives 36.8698…∘36.8698\ldots^\circ.

Keep the same target angle and opposite length, but supply the hypotenuse rather than the adjacent side. This changes the available ratio, not the angle.

Worked example 2: opposite and hypotenuse

Question. A right triangle has opposite side 6 cm6\,\mathrm{cm} and hypotenuse 10 cm10\,\mathrm{cm}, relative to θ\theta. Calculate θ\theta, correct to 11 decimal place.

Plan. Use sine because the known denominator is now the hypotenuse.

sin⁡θ=610opposite divided by hypotenuseθ=sin⁡−1(610)reverse sine=36.8698…∘use degree mode\begin{aligned} \sin\theta&=\frac6{10} && \text{opposite divided by hypotenuse}\\ \theta&=\sin^{-1}\left(\frac6{10}\right) && \text{reverse sine}\\ &=36.8698\ldots^\circ && \text{use degree mode} \end{aligned}

Answer. θ=36.9∘\theta=36.9^\circ.

Check. Pythagoras recovers adjacent side 102−62=8\sqrt{10^2-6^2}=8. Then cos⁡−1(8/10)=36.8698…∘\cos^{-1}(8/10)=36.8698\ldots^\circ, confirming the angle by a different ratio.

Watch out

tan⁡−1(0.75)\tan^{-1}(0.75) does not mean 1/tan⁡(0.75)1/\tan(0.75). The superscript here names an inverse function, which returns an angle. It is not the reciprocal of a tangent value.

Choose a ratio from a fresh pair of lengths.

Finding an acute angle

The ratio must describe the angle asked for. If one shorter side is given with the hypotenuse, decide whether that shorter side is opposite or adjacent to the target. Its length alone cannot tell you which inverse key to press.

Write the ratio equation, apply the matching inverse function, and check that the result lies between 0∘0^\circ and 90∘90^\circ. Within this acute range, each of the three ratios identifies a single angle. The other acute angle is its complement: the two add to 90∘90^\circ.

Worked example 3: the known side is adjacent

Question. A right triangle has hypotenuse 13 cm13\,\mathrm{cm}. The side adjacent to θ\theta is 9 cm9\,\mathrm{cm}. Calculate θ\theta, correct to 11 decimal place.

Plan. Use inverse cosine for adjacent and hypotenuse.

cos⁡θ=913identify the known side’s roleθ=cos⁡−1(913)undo cosine=46.1869…∘evaluate the complete fraction\begin{aligned} \cos\theta&=\frac9{13} && \text{identify the known side's role}\\ \theta&=\cos^{-1}\left(\frac9{13}\right) && \text{undo cosine}\\ &=46.1869\ldots^\circ && \text{evaluate the complete fraction} \end{aligned}

Answer. θ=46.2∘\theta=46.2^\circ.

Check. The opposite side is 132−92=88\sqrt{13^2-9^2}=\sqrt{88}. Thus tan⁡−1(88/9)=46.1869…∘\tan^{-1}(\sqrt{88}/9)=46.1869\ldots^\circ, agreeing with the cosine calculation.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account