Pythagoras and trigonometry in right-angled triangles – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(\sqrt{169-25}=\sqrt{144}=\) 12 cm (1 method, 1 answer) Subtract the square of the known side from the square of the hypotenuse, then take the square root. The other side is 12 cm.
   2. \(\sin x=\dfrac5{13}\) (1), \(x=\) \(22.6^\circ\) (1) The smallest angle is opposite the smallest side, 5 cm. Use sine with the hypotenuse 13 cm, then \(\sin^{-1}\).

   Check (a): 5, 12, 13 is a Pythagorean triple, since \(25+144=169\).

2. Answer to Question 2 (4 marks)

   1. \(7\tan38^\circ=\) 5.47 cm (1 for the equation, 1 for the answer) The adjacent side is known and we need the opposite side, so we use tangent: opposite \(=\) adjacent \(\times\tan38^\circ\).
   2. \(\dfrac7{\cos38^\circ}=\) 8.88 cm (1 for the equation, 1 for the answer) Use cosine: hypotenuse \(=\) adjacent \(\div\cos38^\circ\). Check with Pythagoras: \(\sqrt{7^2+5.47^2}\approx8.88\).

   In (a), the calculator must be in degree mode; in radian mode \(\tan38\) gives a meaningless value. Rounding 5.47 too early would spoil the check in (b).

3. Answer to Question 3 (4 marks)

   1. \(\tan x=\dfrac{18}{30}=0.6\) (1), \(x=\) \(31.0^\circ\) (2) The height of the pole is opposite the angle and the 30 m distance is adjacent, so we use tangent and then \(\tan^{-1}\).
   2. \(\sqrt{900+324}=\) 35.0 m (1) The line from the person to the top is the hypotenuse, so we use Pythagoras with 30 m and 18 m.

   Check (b) with trigonometry: \(30\div\cos31.0^\circ\approx35.0\) m. The exact value is \(\sqrt{1224}\approx34.99\) m. A common error in (a) is to use sine with 18 and 30, treating 30 m as the hypotenuse.

4. Answer to Question 4 (4 marks)

   1. \(\cos60^\circ=\dfrac12\) (1) and \(\tan45^\circ=1\) (1) Read the exact values from the table of special angles. In a right-angled isosceles triangle the opposite and adjacent sides are equal, so \(\tan45^\circ=1\).
   2. \(8\sin30^\circ=8\times\dfrac12=\) 4 cm (2) The side opposite the \(30^\circ\) angle uses sine: opposite \(=\) hypotenuse \(\times\sin30^\circ\).

   In (b), the triangle is half of an equilateral triangle with side 8 cm, which explains why the side opposite \(30^\circ\) is half the hypotenuse. The third side is \(\sqrt{64-16}=4\sqrt3\) cm.

5. Answer to Question 5 (4 marks)

   1. Base diagonal \(=\sqrt{64+36}=10\) (1); space diagonal \(=\sqrt{100+16}=\) 10.77 cm (1) The longest rod is the space diagonal. First find the base diagonal with Pythagoras, then use Pythagoras again with the height 4 cm.
   2. \(\tan x=\dfrac4{10}\) (1), \(21.8^\circ\) (1) The height 4 cm is opposite the angle and the base diagonal 10 cm is adjacent, so we use tangent.

   Check (a) in one step: \(\sqrt{8^2+6^2+4^2}=\sqrt{116}\).

Total : 20 marks