## Answers

Sop? Given that 2016?): ila and TT < 2 3TT 4 2 la) - base base We know that in right angled triangle. cosine of angle hypotenuse.

Ie hypo. Since -T. So, (numerically) Hypotenuse= 4 Now using Pythagoras theorem (Hypo.): bere) + ( perpendicular) Pesrependicular 102-1 VIS cas d base = > Hence the missing side is Vis. We know that cos²o + sinzo = 1 .: cos at sin'd=1 2 el 1 +115 sin $ = "/i-coroa TG 4

c) We have sing ris and cosd a -- 2 Since tamd sind ITS/4 11 cos 2 tom o 2 - Vis ed). sino=2 sino cosa 2 Using identity sino = 2.15 sam we get 27 / DATA » sind - - Sis 8 [:d lies between (e).

Cos od=-11-sin2 -S1-15 π and 3 IT Z Floo 64 Jis ( f ) tan o = sind - codo 7 19) Again sin 20 2 sino coso = 2X-ITS X-I TIS 17 ६५ 8 32

ch) Also cos 20 = 1 + 1-sin 20 4905 [ is in 2p=7115 1024 32 Et 17 32 E tan 26= sin 20 cos20 TVIS .(ta 17 (j). since cos d = w - - 1 2 = 205"' (*) 104.48 (k). Also sin 20 = 7515 32 20= sint (7JTS (75) - 58.21 W Since cos o = -1 $ =.cos! (-1) T 8 . $= 151.04

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