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

1 Pythagorean theorem and converse / 4 pts

1. \(b^2 = 41^2 - 9^2 = 1681 - 81 = 1600\) (1 pt), so \(b = 40\) in (1 pt).
2. The longest side is \(26\): \(10^2 + 24^2 = 100 + 576 = 676\) (1 pt) and \(26^2 = 676\). By the converse, the triangle is right (1 pt).

2 Special right triangles / 3 pts

1. Leg \(= \dfrac{14}{\sqrt{2}} = 7\sqrt{2}\) (1 pt) \(\approx 9.90\) cm (0.5 pt).
2. The long leg is \(s\sqrt{3} = 9\sqrt{3}\), so the short leg is \(s = 9\) ft (1 pt) and the hypotenuse is \(2s = 18\) ft (0.5 pt).

3 Geometric mean / 3 pts

Altitude: \(h^2 = 5 \cdot 20 = 100\), so \(h = 10\) (1 pt). Hypotenuse: \(5 + 20 = 25\) (0.5 pt).

Legs: \(\sqrt{5 \cdot 25} = \sqrt{125} = 5\sqrt{5} \approx 11.18\) and \(\sqrt{20 \cdot 25} = \sqrt{500} = 10\sqrt{5} \approx 22.36\) (1.5 pts).

Check: \(125 + 500 = 625 = 25^2\).

4 Right triangle trigonometry / 4 pts

1. \(\cos 62^\circ = \dfrac{11}{h}\), so \(h = \dfrac{11}{\cos 62^\circ} \approx 23.43\) cm (2 pts). \(\tan 62^\circ = \dfrac{o}{11}\), so \(o = 11\tan 62^\circ \approx 20.69\) cm (1 pt).
2. The smaller angle is opposite the leg \(9\): \(\tan A = \dfrac{9}{40}\), so \(A = \tan^{-1}\!\left(\dfrac{9}{40}\right) \approx 12.7^\circ\) (1 pt).

5 Angle of depression / 3 pts

1. It is also \(25^\circ\): the horizontal line at the balcony and the street are parallel, so the angles are alternate interior angles (1 pt).
2. \(\tan 25^\circ = \dfrac{30}{d}\) (1 pt), so \(d = \dfrac{30}{\tan 25^\circ} \approx 64.3\) m (1 pt).

6 Law of Cosines and Law of Sines / 3 pts

1. \(a^2 = 11^2 + 16^2 - 2(11)(16)\cos 52^\circ = 377 - 352\cos 52^\circ \approx 160.29\) (1 pt), so \(a \approx 12.66\) (1 pt).
2. \(\sin B = \dfrac{11\sin 52^\circ}{12.66} \approx 0.6847\), so \(B \approx 43.2^\circ\) (1 pt). It is acute because \(b < a\).