Finding an Unknown Angle

When two sides of a right-angled triangle are known, write the ratio that uses them and undo it with sin⁻¹, cos⁻¹ or tan⁻¹ on the calculator to find the angle.

Before you start

Choosing the ratio (lessons 10.2 and 10.3): from the angle, name the two sides involved. Opposite and hypotenuse: $\sin$ . Adjacent and hypotenuse: $\cos$ . Opposite and adjacent: $\tan$ .
The two acute angles of a right-angled triangle add up to $90^{\circ}$ . Angles are given correct to 1 decimal place.

Undoing a Ratio

So far we have gone from an angle to its ratio: $\sin 30^{\circ} = 0.5$ . To find an angle we go the other way: from $\sin \theta = 0.5$ back to $\theta = 30^{\circ}$ . The calculator does this with the keys $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ , usually reached by pressing SHIFT and then sin, cos or tan.

The inverse ratios

If $\sin \theta = k$ , then $\theta = \sin^{-1} k$ .

If $\cos \theta = k$ , then $\theta = \cos^{-1} k$ .

If $\tan \theta = k$ , then $\theta = \tan^{-1} k$ .

$\sin^{-1} k$ is read "inverse sine of $k$ ": the angle whose sine is $k$ . The $-1$ is not a power, so $\sin^{-1} 0.5$ is not $\frac{1}{\sin 0.5}$ .

The calculator must be in degree mode. A sine or a cosine is never more than $1$ , so $\sin^{-1} 1.2$ gives an error.

Example 1: An angle from the value of its ratio

Find the acute angle, correct to 1 decimal place where needed: (a) $\sin A = 0.5$ , (b) $\cos B = 0.4$ , (c) $\tan C = 2.5$ .

Solution

(a) $A = \sin^{-1} 0.5$ . Key SHIFT sin 0.5 =: $A = 30^{\circ}$ .

(b) $B = \cos^{-1} 0.4$ . Key SHIFT cos 0.4 =: $66.421\ldots$ , so $B = 66.4^{\circ}$ (1 d.p.).

(a) $A = 30^{\circ}$ (b) $B = 66.4^{\circ}$ (c) $C = 68.2^{\circ}$

An Angle from Two Sides

Finding an angle

1. From the angle you want, name the two sides you know.

2. Choose the ratio that uses those two sides, and write it as a fraction.

3. Key the inverse of the whole fraction in one go, then round to 1 decimal place.

Example 2: Using sine

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $QR = 15$ cm and $PR = 22$ cm. Find $\angle QPR$ , marked $\theta$ .

Solution

From $\theta$ at $P$ : $QR = 15$ cm is opposite and $PR = 22$ cm is the hypotenuse. Opposite and hypotenuse: sine.

$\sin \theta = \dfrac{15}{22}$

$\theta = \sin^{-1}\left(\dfrac{15}{22}\right)$ (key SHIFT sin ( 15 ÷ 22 ) =)

$= 42.985\ldots = 43.0^{\circ}$ (1 d.p.)

$\angle QPR = 43.0^{\circ}$ (to 1 d.p.)

Example 3: Using cosine

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $AC = 13$ cm and $AB = 22$ cm. Find $\angle A$ , marked $\theta$ .

Solution

From $\angle A$ : $AC = 13$ cm is adjacent and $AB = 22$ cm is the hypotenuse. Adjacent and hypotenuse: cosine.

$\cos A = \dfrac{13}{22}$ , so $A = \cos^{-1}\left(\dfrac{13}{22}\right) = 53.778\ldots = 53.8^{\circ}$ (1 d.p.)

$\angle A = 53.8^{\circ}$ (to 1 d.p.)

Example 4: Using tangent

In $\triangle XYZ$ , $\angle Y = 90^{\circ}$ , $XY = 23.6$ cm and $YZ = 15.5$ cm. Find $\angle X$ , marked $\theta$ .

Solution

From $\angle X$ : $YZ = 15.5$ cm is opposite and $XY = 23.6$ cm is adjacent. The hypotenuse is not given. Opposite and adjacent: tangent.

$\tan X = \dfrac{15.5}{23.6}$ , so $X = \tan^{-1}\left(\dfrac{15.5}{23.6}\right) = 33.296\ldots = 33.3^{\circ}$ (1 d.p.)

$\angle X = 33.3^{\circ}$ (to 1 d.p.)

Both Acute Angles

Once one acute angle is found, the other is $90^{\circ}$ minus it. Subtract the full calculator value, not the rounded angle. Or find it with its own ratio, as a check.

Example 5: Choosing the ratio, and the other angle

In $\triangle PQR$ , $\angle R = 90^{\circ}$ , $PQ = 18$ m and $QR = 12$ m. Find $\angle P$ and $\angle Q$ .

Solution

The right angle is at $R$ , so $PQ = 18$ m is the hypotenuse.

From $\angle P$ : $QR$ does not touch $P$ , so it is opposite. Opposite and hypotenuse: sine.

$\sin P = \dfrac{12}{18}$ , so $P = \sin^{-1}\left(\dfrac{12}{18}\right) = 41.810\ldots = 41.8^{\circ}$ (1 d.p.)

$\angle Q = 90^{\circ} - 41.810\ldots^{\circ} = 48.189\ldots = 48.2^{\circ}$ (1 d.p.)

Check: from $\angle Q$ , $QR$ is adjacent, and $\cos^{-1}\left(\dfrac{12}{18}\right) = 48.189\ldots^{\circ}\text{,}$ the same.

$\angle P = 41.8^{\circ}$ and $\angle Q = 48.2^{\circ}$ (to 1 d.p.)

Angles in Settings

In a setting, draw the right-angled triangle and mark the angle asked for: an angle with the ground is at the foot, an angle with a wall is at the wall. Once the angle is known, it can be used to find a side.

Example 6: A ramp, its angle and then a side

A straight ramp $2.5$ m long rises $0.4$ m from the level ground. (a) Find the angle the ramp makes with the ground. (b) Use the angle to find the horizontal distance the ramp covers, and check it by Pythagoras' theorem.

Solution

(a) From $\theta$ at $F$ : the rise $BT = 0.4$ m is opposite and the ramp $FT = 2.5$ m is the hypotenuse: sine.

$\sin \theta = \dfrac{0.4}{2.5} = 0.16\text{,}$ so $\theta = \sin^{-1} 0.16 = 9.206\ldots = 9.2^{\circ}$ (1 d.p.)

(b) $FB$ is adjacent: $d = 2.5 \cos 9.206\ldots^{\circ} = 2.467\ldots = 2.47$ (3 s.f.), keying the angle from the calculator's memory (Ans), not $9.2$ .

Check: $d = \sqrt{2.5^2 - 0.4^2} = \sqrt{6.09} = 2.467\ldots\text{,}$ the same.

(a) $9.2^{\circ}$ (b) $2.47$ m

Example 7: A plank over a wall

A plank $4$ m long rests on top of a vertical wall $1.8$ m high, with its lower end on the level ground and $1.2$ m of the plank beyond the top of the wall. Find the angle the plank makes with the wall.

Solution

The plank from the ground to the top of the wall is $GT = 4 - 1.2 = 2.8$ m. Only this part is in the right-angled triangle $GWT$ .

The wall stands on level ground, so $\angle W = 90^{\circ}$ and $GT$ is the hypotenuse.

The angle with the wall is at $T$ . From $\theta$ at $T$ : the wall $WT = 1.8$ m is adjacent. Adjacent and hypotenuse: cosine.

$\cos \theta = \dfrac{1.8}{2.8}$ , so $\theta = \cos^{-1}\left(\dfrac{1.8}{2.8}\right) = 49.994\ldots = 50.0^{\circ}$ (1 d.p.)

The plank makes an angle of $50.0^{\circ}$ with the wall.

Isosceles Triangles

An isosceles triangle need not have a right angle, but the line from its apex to the middle of its base is always perpendicular to the base. It cuts the triangle into two equal right-angled triangles, each with half the base.

Example 8: The angles of an isosceles triangle

In $\triangle ABC$ , $AB = AC = 10$ cm and $BC = 14$ cm. Find $\angle ABC$ and $\angle BAC$ .

Solution

Let $M$ be the midpoint of $BC$ . Then $AM$ is perpendicular to $BC$ and $BM = 14 \div 2 = 7$ cm.

In $\triangle ABM$ , $\angle AMB = 90^{\circ}$ . From $\angle B$ : $BM = 7$ cm is adjacent and $AB = 10$ cm is the hypotenuse: cosine.

$\cos B = \dfrac{7}{10}$ , so $\angle ABC = \cos^{-1} 0.7 = 45.572\ldots = 45.6^{\circ}$ (1 d.p.)

The base angles are equal, so $\angle BAC = 180^{\circ} - 2 \times 45.572\ldots^{\circ} = 88.854\ldots = 88.9^{\circ}$ (1 d.p.)

$\angle ABC = 45.6^{\circ}$ and $\angle BAC = 88.9^{\circ}$ (to 1 d.p.)

Common Mistakes

Writing " $\sin \theta = 0.5$ , so $\theta = 0.5^{\circ}$ ". The angle is $\sin^{-1} 0.5 = 30^{\circ}$ .
Radian mode: $\sin^{-1} 0.5$ then shows $0.523\ldots$ instead of $30$ .
Turning the ratio upside down. With opposite $5$ and adjacent $8$ , $\tan \theta = \frac{5}{8}$ and $\theta = 32.0^{\circ}$ ; $\tan^{-1}\left(\frac{8}{5}\right) = 58.0^{\circ}$ is the other acute angle.
Rounding the ratio before the inverse. $\frac{15}{22}$ rounded to $0.68$ gives $\sin^{-1} 0.68 = 42.8^{\circ}\text{,}$ not $43.0^{\circ}$ . Key the fraction.
Finding the angle at the wrong corner: an angle with a wall is where the ladder or plank meets the wall, not at the ground.

Practice Questions

Give angles correct to 1 decimal place and lengths to 3 significant figures. Try each question before opening its solution.

Question 1

Find the acute angle: (a) $\sin A = 0.78$ , (b) $\cos B = 0.35$ , (c) $\tan C = 0.9$ .

ShowHide solution

(a) $A = \sin^{-1} 0.78 = 51.260\ldots$

(b) $B = \cos^{-1} 0.35 = 69.512\ldots$

(a) $51.3^{\circ}$ (b) $69.5^{\circ}$ (c) $42.0^{\circ}$

Question 2

A right-angled triangle has a base of $8.7$ cm and a height of $5.7$ cm. Find the angle $\theta$ between the base and the hypotenuse.

ShowHide solution

From $\theta$ : the height $5.7$ cm is opposite and the base $8.7$ cm is adjacent: tangent.

$\theta = \tan^{-1}\left(\dfrac{5.7}{8.7}\right) = 33.231\ldots$

$33.2^{\circ}$

Question 3

In $\triangle KLM$ , $\angle M = 90^{\circ}$ , $KL = 9.4$ cm and $KM = 6.1$ cm. Find $\angle K$ and $\angle L$ .

ShowHide solution

From $\angle K$ : $KM$ is adjacent and $KL$ the hypotenuse: $\cos K = \dfrac{6.1}{9.4}$ , $K = 49.538\ldots = 49.5^{\circ}\text{.}$

$\angle L = 90^{\circ} - 49.538\ldots^{\circ} = 40.461\ldots = 40.5^{\circ}\text{.}$ (Check: $\sin^{-1}\left(\frac{6.1}{9.4}\right)$ gives the same.)

$\angle K = 49.5^{\circ}$ , $\angle L = 40.5^{\circ}$

Question 4

In $\triangle DEF$ , $\angle F = 90^{\circ}$ , $DE = 13$ cm and $EF = 5$ cm. Find $\angle EDF$ .

ShowHide solution

From $\angle D$ : $EF$ does not touch $D$ , so it is opposite; $DE$ is the hypotenuse: sine.

$\angle EDF = \sin^{-1}\left(\dfrac{5}{13}\right) = 22.619\ldots$

$\angle EDF = 22.6^{\circ}$

Question 5

A straight road rises $1$ m for every $8$ m measured along the level. Find the angle the road makes with the horizontal.

ShowHide solution

In the right-angled triangle, the rise $1$ m is opposite the angle and the level distance $8$ m is adjacent: tangent.

$\theta = \tan^{-1}\left(\dfrac{1}{8}\right) = 7.125\ldots$

$7.1^{\circ}$

Question 6

A ladder $4$ m long leans against a vertical wall, with its foot on level ground $1.2$ m from the wall. Find the angle the ladder makes (a) with the ground, (b) with the wall.

ShowHide solution

(a) The angle with the ground, $\alpha$ , is at $F$ . $FW = 1.2$ m is adjacent and the ladder is the hypotenuse: $\alpha = \cos^{-1}\left(\dfrac{1.2}{4}\right) = 72.542\ldots = 72.5^{\circ}\text{.}$

(b) The angle with the wall, $\beta$ , is at $T$ : $\beta = 90^{\circ} - 72.542\ldots^{\circ} = 17.457\ldots = 17.5^{\circ}\text{.}$

(a) $72.5^{\circ}$ (b) $17.5^{\circ}$

Question 7

A rectangular field $ABCD$ is $80$ m long and $45$ m wide. A straight path runs across it from corner $A$ to corner $C$ . Find the angle $\theta$ between the path and the longer side $AB$ .

ShowHide solution

In $\triangle ABC$ , $\angle B = 90^{\circ}$ . From $\theta$ at $A$ : $BC = 45$ m is opposite and $AB = 80$ m adjacent: tangent.

$\theta = \tan^{-1}\left(\dfrac{45}{80}\right) = 29.357\ldots$

$29.4^{\circ}$

Question 8

The frame of a roof is an isosceles triangle with two sloping sides of $7$ m each and a horizontal base of $12$ m. Find the angle each sloping side makes with the base, and the angle between a sloping side and the upright line from the top to the middle of the base.

ShowHide solution

The upright line from the top to the middle of the base is perpendicular to the base and cuts it into two halves of $6$ m.

In each right-angled triangle, from the base angle: $6$ m is adjacent and $7$ m is the hypotenuse.

Base angle $= \cos^{-1}\left(\dfrac{6}{7}\right) = 31.002\ldots = 31.0^{\circ}$ (1 d.p.)

From the angle at the top, between the sloping side and the upright line: $6$ m is opposite and $7$ m is the hypotenuse, so the angle $= \sin^{-1}\left(\dfrac{6}{7}\right) = 58.997\ldots = 59.0^{\circ}$ (1 d.p.)

Check: the two acute angles of the right-angled triangle add up to $90^{\circ}$ , and $90^{\circ} - 31.002\ldots^{\circ} = 58.997\ldots^{\circ}\text{.}$

$31.0^{\circ}$ with the base; $59.0^{\circ}$ with the upright line

Question 9

Find the error in each. (a) Mei Ling writes " $\cos \theta = 0.5$ , so $\theta = 0.5^{\circ}$ ". (b) In a right-angled triangle the side opposite $\theta$ is $7$ cm and the side adjacent to it is $12$ cm; Kumar writes " $\tan \theta = \frac{12}{7}$ , so $\theta = 59.7^{\circ}$ ".

ShowHide solution

(a) $0.5$ is the cosine, not the angle. $\theta = \cos^{-1} 0.5 = 60^{\circ}\text{.}$

(b) Tangent is opposite over adjacent: $\tan \theta = \frac{7}{12}$ and $\theta = \tan^{-1}\left(\frac{7}{12}\right) = 30.256\ldots = 30.3^{\circ}\text{.}$ His $59.7^{\circ}$ is the other acute angle.

(a) $\theta = 60^{\circ}$ (b) $\theta = 30.3^{\circ}$

Question 10

For Example 4, Ahmad writes $\tan \theta = \frac{15.5}{23.6} = 0.66\text{,}$ then keys $\tan^{-1} 0.66$ and gives $\theta = 33.4^{\circ}$ . Explain why his answer differs from $33.3^{\circ}$ .

ShowHide solution

$\frac{15.5}{23.6} = 0.6567\ldots\text{.}$ Rounding it to $0.66$ changed the angle: $\tan^{-1} 0.66 = 33.424\ldots\text{,}$ while $\tan^{-1} 0.6567\ldots = 33.296\ldots$

Key $\tan^{-1}(15.5 \div 23.6)$ in one go and round only the angle.

He rounded the ratio first; $\theta = 33.3^{\circ}$ .

Question 11

In $\triangle ABC$ , $\angle ABC = 90^{\circ}$ , $AB = 9$ cm and $BC = 12$ cm. $D$ is the point on $BC$ such that $\angle BAD = 25^{\circ}$ . Find (a) $\angle BAC$ , (b) $BD$ , (c) $\angle DAC$ .

ShowHide solution

(a) From $\angle BAC$ : $BC$ is opposite and $AB$ adjacent: $\angle BAC = \tan^{-1}\left(\dfrac{12}{9}\right) = 53.130\ldots = 53.1^{\circ}$

(b) In $\triangle ABD$ , from $\angle BAD$ : $BD$ is opposite and $AB$ adjacent: $BD = 9 \tan 25^{\circ} = 4.196\ldots = 4.20$ cm (3 s.f.)

(a) $53.1^{\circ}$ (b) $4.20$ cm (c) $28.1^{\circ}$

Key Points

$\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ undo the ratios: if $\sin \theta = k$ , then $\theta = \sin^{-1} k$ . The $-1$ is not a power.
From the angle you want, name the two known sides, choose the ratio, and write it as a fraction.
Key the inverse of the whole fraction in degree mode, and round the angle to 1 decimal place at the end.
The other acute angle is $90^{\circ}$ minus the first; subtract the full value.
In an isosceles triangle, the line from the apex to the middle of the base makes two right-angled triangles.