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Unformatted text preview: Maths Study Sheet 18 Introduction Twe useful
geometrical
results angles 8: sides of a triangle part 1
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII This UNIT censists ef 2 parts Part 1: Yeu can calculate the Luiknewn angles and sides in a
rightvangled triangle. When yeu have werked threugh Part 1, yeu sheuld be able te I label a triangle with apprepriate netatien
I use the fact that the angles ef a triangle add up te 18D“ I use Pythageras’ theeretn I cheese an apprepriate trigenernetrical ratie frern sine, cesine er tangent in
erder te calculate a specified side er angle in a rightangled triangle. Prerequisites:
Yeu sheuld be familiar with the material ceirered in the UNITS en I angles limit 14 Part 1] I sine. cesine and tangent [unit 15] Yeti will need a calculater with the scientific functiens SIN. C03 and TAN. Netatien: 'l‘e laleel a triangle we use uppercase letters fer the vertices
{angles} and iewercase letters fer the lengths ef the sides. B A typical rightangled triangle ABC might have angle C = 90“. The cerwentien is te label the
side eppesite angle A as it Then BC er a stand fer
” the length uf side BC” Similarly the side eppesite angle B is h A b C The liypeteu use is eppesite the rightangle C and has length r. l The angles in any triangle add up te 180 degrees.
In a rightangled triangle, since C = '90:“. then
A + B = 90d. Fer example. if A = T2“
then 72” + B = 9U”
B 2 9GB —72"“ =18“. II A wellsknewn theerem named after the mathematician Pythageras
gives a cennectien between the sides ef a rightangled triangle.
It says:
In a rightangled triangle, the square en the hypetenuse is equal te the
sum ef the squares en the ether twe sides.
In eur netatien, this is c2 = aI + £11 Example 1 If a = 5 and i: = 12
then c2 = 5? + 122
: 25 + 144
: 1sa___
c a ﬂea = 13
Example 2 If c = 1? and a = B
then 1?2 z 33 + h?
1:” — as” = a:
289 — 64 : b2
225 = £11
+. a = «Jig = 15. New when we knew either (i) the lengths ef twe sides
er {ii} the size ef ene angle and the length ef ene side it is pessible te use the twe geemetrical results tegether with the
trigenemetrical raties sine, cesine and tangent te find all the missing sides and angles ef the right—angled triangle.
Mathematicians call this precess ”selving the triangle”.
'I‘ype {i} When the lengths et twe sides are knewn
Step (a) use Pythageras’ theerem te find the third side Step {b} use an apprepriate ratie chesen frem sine, cesine er
tangent te find ene ef the angles. Step {c} use ”angles ef a triangle add up te 180” “ te find the third
angle.
Type {ii} When ene angle and one side are knewn Step (a) use the apprepriate ratie chesen fretn sine, cesine,
tangent te find anether side. Step {[3} use Pythageras’ theerem te find the 3rd side. Step (c) use ”angles ef a triangle add up te 180‘“ “
te find the 3rd angle. Example 3 Given r = 13, lJ' : 5 find the other sides and angles of the triangle. This is a type {i} example. Ste p {a} We use Pythagoras“ theorem to find length it
(see example 1 above) e212 5132]} (bl TU find angle A l3
The given information tells us the ‘3
lengths of the hypotenuse and the adjacent side to angle A.
So we use the cosine ratio. . A I cosrl = 1:5 — 0.3846 A 5 C A = cos'1 [1.3846 Ior arc cos 0.3846] [Use your calculator in degree mode with keys INV COS or ARC COS] A = 6?.4”
Step (ch B = 90” — 67.4:ﬂ
E : 22.6” Example 4 rGiven B — 58“ and b = 7.5 find the other sides and angles of the triangle. This is a type {ii} example. Step (a) Suppose we decide to find ﬂ'lE! length of side c. Then we want the ratio that relates the hypotenuse and opposite
side to angle B, i.e. the sine ratio. sin B : 7' 5
c
Taking reciprocals,
1 _s_
sin B 5
Multiplying by T5.
7.5
. = c
Sill B
7. 5 7’ 5 E: __ : ———' =3344 3d. .
sin 58“ 0.843 ( P) Step {13} Use Pythageraa’ theerem
{3.844)2 = {3315}? + bi
76.216: 56.25 + b2 T6216 — 56.25 £72
21.966: if
i? = 4.63? 90:” — 58°
32“ Step (e) Emil
ll Alternatively, in step (a) we might have ehesen in find length I! first tan 53 . = E
b 1 : i [an 58” 7.5
A .
tan 58"
ii = 4.631 etc.
Exercise In triangle ABC, which is right—angled at C 1. it a : 4, C = 5 find angle A and length b.
2. if B = 62”, a = 8.5 find lengths in and c.
3. if e = 25, a = 13 Find angle B. 4. if B = '30” and a = 12.5 find the ether sides; and angle. Solutions 1. 14:53.1” h  
DJ 2. b = 8.5 tan 62“ _ 16
r: = 18.1 i 3. ii“ = 43.9“ 4. i1 = 6.25 , e = 10.8 . A = n‘fer‘J‘:i ...
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 Spring '11
 DR.MCDONALD

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