Right Triangles and Trigonometry: practice solutions, Grade 10 – download the PDF

1 Zip line cable ★★★

The pole is vertical and the ground is level, so the triangle is right and the cable is the hypotenuse.

\(c^2 = 12^2 + 16^2 = 144 + 256 = 400\), so \(c = 20\).

The cable is 20 m long.

2 Missing leg ★★★

Use \(a^2 + b^2 = c^2\) with \(c = 26\) and \(a = 10\): \(b^2 = 26^2 - 10^2 = 676 - 100 = 576\).

So \(b = \sqrt{576} = 24\). The other leg is 24 cm.

3 Is it a right triangle? ★★★

1. The longest side is \(17\): \(8^2 + 15^2 = 64 + 225 = 289 = 17^2\). By the converse, the triangle is right.
2. The longest side is \(10\): \(6^2 + 7^2 = 36 + 49 = 85 < 100 = 10^2\). The triangle is obtuse, so it is not right.

4 Isosceles right triangle ★★★

1. The hypotenuse is leg \(\times \sqrt{2}\): \(6\sqrt{2} \approx 8.49\) in.
2. Each leg is \(\dfrac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07\) in.

5 Half an equilateral triangle ★★★

The short leg is \(s = 5\). The long leg is \(s\sqrt{3} = 5\sqrt{3} \approx 8.66\) cm and the hypotenuse is \(2s = 10\) cm.

6 Naming the ratios ★★★

For \(A\): opposite \(= 5\), adjacent \(= 12\), hypotenuse \(= 13\). \(\sin A = \dfrac{5}{13}\), \(\cos A = \dfrac{12}{13}\), \(\tan A = \dfrac{5}{12}\).

For \(B\) the roles swap: \(\sin B = \dfrac{12}{13}\), \(\cos B = \dfrac{5}{13}\), \(\tan B = \dfrac{12}{5}\). Notice \(\sin A = \cos B\), as expected for complementary angles.

7 True or false? ★★★

1. True. \(35^\circ\) and \(55^\circ\) are complementary, and the sine of an angle equals the cosine of its complement.
2. False. The opposite leg is shorter than the hypotenuse, so \(\sin A < 1\).
3. True. A tangent is a ratio of two legs and can be any positive number; here the opposite leg is three times the adjacent leg (about \(71.6^\circ\)).
4. False. \(\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}\); the ratio opposite/adjacent is the tangent.

8 Geometric means ★★★

1. \(\sqrt{4 \cdot 25} = \sqrt{100} = 10\).
2. \(\sqrt{6 \cdot 8} = \sqrt{48} = 4\sqrt{3} \approx 6.93\).

9 Altitude to the hypotenuse ★★★

\(CD^2 = AD \cdot DB = 36\), so \(CD = 6\). \(AB = 3 + 12 = 15\).

\(AC^2 = AD \cdot AB = 3 \cdot 15 = 45\), so \(AC = 3\sqrt{5} \approx 6.71\). \(BC^2 = DB \cdot AB = 12 \cdot 15 = 180\), so \(BC = 6\sqrt{5} \approx 13.42\).

Check: \(45 + 180 = 225 = 15^2\).

10 Water slide ★★★

The slide is the hypotenuse. Height (opposite): \(18\sin 35^\circ \approx 10.32\) m. Horizontal length (adjacent): \(18\cos 35^\circ \approx 14.74\) m.

Check: \(10.32^2 + 14.74^2 \approx 106.5 + 217.3 = 323.8 \approx 18^2 = 324\).

11 Angles of a right triangle ★★★

The hypotenuse is \(\sqrt{64 + 225} = 17\). The angle \(A\) opposite the leg \(8\) satisfies \(\tan A = \dfrac{8}{15}\), so \(A = \tan^{-1}\!\left(\dfrac{8}{15}\right) \approx 28.1^\circ\).

The other angle is \(90^\circ - 28.1^\circ \approx 61.9^\circ\).

12 Wheelchair ramp ★★★

\(\tan A = \dfrac{2.5}{30}\), so \(A = \tan^{-1}\!\left(\dfrac{2.5}{30}\right) \approx 4.8^\circ\).

Length: \(\sqrt{30^2 + 2.5^2} = \sqrt{906.25} \approx 30.10\) ft.

The ramp makes about a 4.8° angle and is about 30.1 ft long.

13 Height of a tower ★★★

The side opposite the angle is \(x = 40\tan 52^\circ \approx 51.2\) m. Adding the eye height, \(51.2 + 1.6 = 52.8\).

The tower is about 52.8 m tall.

14 Lighthouse and boat ★★★

The angle of depression equals the angle of elevation from the boat, \(18^\circ\). In the right triangle, \(\tan 18^\circ = \dfrac{45}{d}\), so \(d = \dfrac{45}{\tan 18^\circ} \approx 138.5\).

The boat is about 138 m from the base. (A common mistake is using \(45\tan 18^\circ\).)

15 Law of Sines ★★★

\(C = 180^\circ - 48^\circ - 65^\circ = 67^\circ\).

\(b = \dfrac{12\sin 65^\circ}{\sin 48^\circ} \approx 14.63\) and \(c = \dfrac{12\sin 67^\circ}{\sin 48^\circ} \approx 14.86\).

16 Law of Cosines ★★★

Two sides and the included angle: use the Law of Cosines. \(d^2 = 9^2 + 14^2 - 2(9)(14)\cos 38^\circ = 277 - 252\cos 38^\circ \approx 78.42\).

So \(d \approx 8.86\). The ends are about 8.86 mi apart.

17 Which law? ★★★

1. Law of Cosines (SSS). In fact \(\cos C = \dfrac{81 + 144 - 225}{216} = 0\), so \(C = 90^\circ\) and the triangle is right.
2. Law of Sines (AAS): \(a = \dfrac{14\sin 35^\circ}{\sin 80^\circ}\).
3. Law of Cosines (SAS: the angle is between the two known sides).
4. Law of Sines (SSA): \(\sin B = \dfrac{28\sin 30^\circ}{20}\), and the ambiguous case must be checked.

18 Planes leaving an airport ★★★

After \(2\) h they have flown \(600\) mi and \(480\) mi. The included angle is \(70^\circ\), so the Law of Cosines applies:

\(d^2 = 600^2 + 480^2 - 2(600)(480)\cos 70^\circ \approx 590{,}400 - 576{,}000(0.3420) \approx 393{,}400\).

\(d \approx 627\). The planes are about 627 mi apart.

19 Two angles of elevation ★★★

Let \(h\) be the height. Then the distance from \(P\) to the base is \(\dfrac{h}{\tan 28^\circ}\) and from \(Q\) to the base is \(\dfrac{h}{\tan 41^\circ}\).

The difference is \(60\): \(h\left(\dfrac{1}{\tan 28^\circ} - \dfrac{1}{\tan 41^\circ}\right) = 60\).

\(h \approx \dfrac{60}{1.8807 - 1.1504} \approx 82.2\). The building is about 82.2 ft tall.

Check: \(82.2/\tan 28^\circ \approx 154.6\) and \(82.2/\tan 41^\circ \approx 94.6\), and \(154.6 - 94.6 = 60\).

20 Special triangles with algebra ★★★

1. The height cuts the triangle into two 30°-60°-90° triangles with short leg \(6\), so the height is \(6\sqrt{3} \approx 10.39\) cm. Area \(= \dfrac{1}{2}(12)(6\sqrt{3}) = 36\sqrt{3} \approx 62.35\) cm\(^2\).
2. The diagonal is \(s\sqrt{2} = 20\), so \(s = 10\sqrt{2} \approx 14.14\) in. Area \(= s^2 = 200\) in\(^2\).
3. Long leg \(= s\sqrt{3} = 9\), so \(s = \dfrac{9}{\sqrt{3}} = 3\sqrt{3} \approx 5.20\) ft. The hypotenuse is \(2s = 6\sqrt{3} \approx 10.39\) ft.

21 The ambiguous case ★★★

By the Law of Sines, \(\sin B = \dfrac{14\sin 30^\circ}{10} = 0.7\). So \(B \approx 44.4^\circ\) or \(B' = 180^\circ - 44.4^\circ \approx 135.6^\circ\).

Both fit: \(30^\circ + 135.6^\circ = 165.6^\circ < 180^\circ\). So there are two triangles.

• Triangle 1: \(C \approx 105.6^\circ\), \(c = \dfrac{10\sin 105.6^\circ}{\sin 30^\circ} \approx 19.3\).
• Triangle 2: \(C' \approx 14.4^\circ\), \(c' = \dfrac{10\sin 14.4^\circ}{\sin 30^\circ} \approx 5.0\).

22 Proving an identity ★★★

1. \(\sin\theta = \dfrac{a}{c}\) and \(\cos\theta = \dfrac{b}{c}\), so \(\sin^2\theta + \cos^2\theta = \dfrac{a^2 + b^2}{c^2} = \dfrac{c^2}{c^2} = 1\) by the Pythagorean theorem.
2. \(\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}\), so \(\cos\theta = \dfrac{4}{5}\) (positive since \(\theta\) is acute). Then \(\tan\theta = \dfrac{3/5}{4/5} = \dfrac{3}{4}\).
3. Draw a triangle with opposite \(5\) and adjacent \(12\). The hypotenuse is \(\sqrt{25 + 144} = 13\), so \(\sin\theta = \dfrac{5}{13}\) and \(\cos\theta = \dfrac{12}{13}\).

23 A triangular lot ★★★

Test the converse: \(120^2 + 150^2 = 14{,}400 + 22{,}500 = 36{,}900 < 40{,}000 = 200^2\). The lot is not right; it is obtuse.

Largest angle \(C\) (opposite \(200\)): \(\cos C = \dfrac{14{,}400 + 22{,}500 - 40{,}000}{2(120)(150)} = \dfrac{-3100}{36{,}000}\), so \(C \approx 94.94^\circ\).

Angle \(A\) (opposite \(120\)): \(\cos A = \dfrac{150^2 + 200^2 - 120^2}{2(150)(200)} = \dfrac{48{,}100}{60{,}000} \approx 0.8017\), so \(A \approx 36.71^\circ\). Angle \(B\) (opposite \(150\)): \(B \approx 48.35^\circ\).

Sum: \(94.94 + 36.71 + 48.35 = 180.00^\circ\).