Eamination Question 1: alculate the angle marked. Give your answer correct to one decimal place. 12 m 15m Eamination Question 2: 20 m Work out the siz

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1 Revision Topic 18: Trigonometry Trigonometry connects the length of sides and angles in right-angled triangles. Some important terms In a right-angled triangle, the side opposite the right angle is called the hypotenuse. The hypotenuse is the longest side in a right-angled triangle. The side opposite the angle of interest is called the opposite. The third side (which is net to the angle of interest) is called the adjacent. pposite ypotenuse djacent Sin, cos and tan Three important formulae connect the lengths of, and with the angle : sin cos tan. You need to learn these formulae! It sometimes helps to remember them as ST. Finding an angle Trigonometry can be used to find angles in right-angled triangles. Eample 1: 14 cm 10 cm Start by labelling the triangle, and. We then have to decide whether to use sin, cos or tan ST Since we know the values of and, we use cos: cos 10 cos To find the angle, you have to press SIFT cos You get: = 44.4 Note: Write down several calculator digits in your working. Eample 2: 20 cm 9cm We start by labelling the triangle. This time we will need to use tan as we know and : ST tan 9 tan To find angle, you press SIFT tan You get: = Eample 3: 11.2m θ 7.3 m The triangle is labelled, and. ecause we know and, we need to use sin this time: ST So, sin 7.3 sin To recover angle θ, you press SIFT sin The answer is: θ = 40.7.

2 Eamination Question 1: alculate the angle marked. Give your answer correct to one decimal place. 12 m 15m Eamination Question 2: 20 m Work out the size of the angle marked. 30 m Finding a side Trigonometry can also be used to find sides in right-angled triangles if you know one of the angles. Eample 1: cm 24 cm 37 Start by labelling the triangle, and. We then have to decide whether to use sin, cos or tan ST Since we want to find and we know, we use sin: sin 37 sin So, = = 14.4 cm. Remember to put a unit on the answer!! Eample 2: 6.8 cm 65 cm We start by labelling the triangle. This time we will need to use tan as we know and want : ST tan 65 tan So, = = 14.6 cm. Eample 3: 0.85 m 52 y The triangle is labelled, and. ecause we know and want, we need to use cos this time: ST So, cos52 y cos y So, y = = m. Dr Duncombe Easter

3 Eamination Question 3: R ngle Q = 90 degrees, angle P= 32 degrees and PR = 2.6 m. 2.6 cm alculate the length of QR. Give your answer in metres to 3 significant figures. P 32 Q Eamination Question 4: Triangle is shown on the right. a) alculate the length of the side. b) alculate the length of the side. 69 6cm Finding the length of sides (continued) Some side lengths are harder to calculate they appear on the denominator of the fractions. onsider the following eample: Eample: Find the length cm in this triangle: 40 ST 18cm cm The triangle is labelled, and as normal. Since we know and want to find we use cos. 18 cos 40 so cos 40 so The side we wish to find appears on the denominator. There are two methods that we could use to find the value of : Method 1: Write both sides as fractions: Invert both sides: Work out the left hand side: = Method 2: Rearrange the equation by first getting rid of the fraction (multiply by ): = 18 Get on its own by dividing by : cm Therefore = = 23.5 cm We can check our answer since we are finding the hypotenuse, it should be the longest side in the triangle. Dr Duncombe Easter

4 Eample 2: 12.5cm pp yp 39 dj alculate the length of. We need to use tan: ST pp tan39 dj 12.5 tan s appears on the bottom, we need to use one of the methods used in the previous eample. Eg: So, cm Eamination Question 5: The triangle has a right angle at. 8.3 cm ngle = 50 degrees and = 8.3 cm. alculate the length of. 50 Eamination Question 6: 6cm 8cm 40 D D is a quadrilateral. ngle D = 90, angle D = 90, angle D = 40. = 6 cm, D = 8cm. a) alculate the length of D. Give your answer correct to 3 significant figures. [int: use Pythagoras theorem!] b) alculate the size of angle D. Give your answer correct to 3 significant figures. c) alculate the length of. Give your answer correct to 3 significant figures. Dr Duncombe Easter

5 Trigonometry: arder Questions: Eample: 16.4 cm D 19.5 cm 19.5 cm = 19.5 cm, = 19.5 cm and = 16.4 cm. ngle D = 90 degrees. D is a straight line. alculate the size of angle. Give your answer correct to 1 decimal place. ecause triangle is isosceles we can just consider the top triangle D. Length D would be half of 16.4 cm, i.e. 8.2 cm. 8.2 cm 19.5 cm cos 8.2 cos D 19.5 SIFTcos Eamination Question 7: D 148 cm and are two sides of a rectangle. = 120 cm and = 148 cm. D is a point on. ngle D = 15 degrees. Work out the length of D. Give your answer to the nearest centimetre cm Dr Duncombe Easter

6 Trigonometry and earings Recall that bearings measure direction. They are angles that are measured clockwise from a north line. earings have three digits. Eample: The diagram shows the path of a jet-ski from P to Q to R. N N 070 P 700m Q m Q is 700m from P on a bearing of 070 degrees. R is 500m from Q on a bearing of 160. alculate the bearing of P from R. R First it is important to realise that triangle PQR is a right-angled triangle (with the right angle at Q): Q m m The blue angles are alternate (Z angles) so are equal. The red angles add up to 180 degrees as they make a straight line. Therefore angle Q is a right angle. 20 P R We can therefore find angle using trigonometry: 700 tan SIFT tan The angle of P from R is the clockwise angle measured from R to P. It is = So the bearing is 286 (to nearest degree). Eamination Question allymena Larne N 32 km 15 km allymena is due west of Larne. Woodburn is 15 km due south of Larne. allymena is 32 km from Woodburn. a) alculate the distance of Larne from allymena, correct to 1 decimal place. b) alculate the bearing of allymena from Woodburn. Woodburn Dr Duncombe Easter

7 ngle of elevation Worked eamination question bbi is standing on level ground, at, a distance of 19 metres away from the foot E of a tree TE. She measures the angle of elevation of the top of the tree at a height of 1.55 metres above the ground as 32. alculate the height TE of the tree. Give your answer correct to 3 significant figures. Solution: First find the length TR using trigonometry, specifically using tan: TR tan TR TR metres So, to get the height of the tree you need to add bbi s height to this distance. Therefore, height of tree = = 13.4 metres (to 3 s.f.) Dr Duncombe Easter

SOLVING RIGHT-ANGLED TRIANGLES

Mathematics Revision Guides Right-Angled Triangles Page 1 of 12 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier SOLVING RIGHT-ANGLED TRIANGLES Version: 2.2 Date: 21-04-2013 Mathematics