u cot u o 2nd option 2 cot u tan u so 5th option 2 sec u csc u so 3rd option 2 csc u sec u so 6th option 2 sin u cos u so 4th option 2 cos u sin u so 1st option 2 3)sin u v sin u cos v cos u sin v...5th option 2) tan cos u v cos u cos v sin u sin v....2nd option cos u v cos u cos v sin u sin v.....4th option sin u v sin u cos v cos u sin v....1st option tan u tan v ....3rd option 1 tan u tan v tan u tan v tan u v 6th option 1 tan u tan v tan u v a c 0.7 c sin sin sin 30 sin 90 0.7 0.7 1 c 1.4, as sin 90 1,sin 30 sin 30 1 / 2 2 5)given that 45, b 8.2, 90 Since the sum of angles of a triangle is 180 so 180 45 90 180 45 b a 8.2 a Then by sine rule we have a 8.2 sin sin sin 45 sin 45 6)given that a 29, c 96, 90 a c 29 96 29 By sine rule we have sin 17.5828 sin sin sin sin 90 96 Since the sum of angles of a triangle is 180 so 180 17.5828 90 180 72.4172 b c b 96 Then by sine rule we have b 96sin 72.4172 91.5150 sin sin sin 72.4172 sin 90 4)By sine rule in triangle ABC we have 7)given that 75.4, c 26, 90 Since the sum of angles of a triangle is 180 so 180 75.4 90 180 14.6 a c a 26 Then by sine rule we have a 6.5538 sin sin sin14.6 sin 90 8)Given Jack height is 6.5 ft. Distance between jack and pole is 60 ft and length is shadow is 24 ft. Let height of street light is x ft. In similar triangles ABC and ADF we have 84 6.5 BC DF 6.5 x x 22.75 23 ft AB AF 24 24 60 24 11) x a cot for 0 , a 0 a a a a So , as a>0 and csc 2 1 cot 2 2 2 2 2 2 2 a x a a cot a 2 1 cot 2 a csc 1 sin so 2nd option is correct csc 1 1 x 116 45, as cos 45 2 2 12) 2 cos x 116 1 cos x 116 x 161 in interval 90,180 13)10sin 2 t 9sin t 7 0 sin t 9 81 280 9 19 1 7 , 2 10 20 2 5 Since 1 sin t 1 for all t so sin t 7 is not possible so discard this value 5 1 t 150 2 Hence sin t 14)sin 1 cos sin cos 1 1 2 sin 1 2 cos 1 2 sin cos 45 cos sin 45 1 2 1 45 45,135 0,90 2 Thus 2nd option is correct sin 45 2 2 ,cos 225 cos 180 45 cos 45 2 2 sin285 sin 60 225 sin 60 cos 225 sin 225 cos 60 15)Note that sin 225 sin 180 45 sin 45 2 1 6 2 so 1st option 2 2 4 16)csc0.24 csc 1.33079 sec1.33079 sec1.33 2 3 2 2 2 3 sin cos so 2nd option 8 8 2 8 18)cos21 sin 5 sin 21 cos5 cos 21 sin 5 sin 21 cos 5 , as cos x cos x 17)sin sin 5 21 , use sin a b sin a cos b cos a sin b sin16 hence 3rd option 19)Given , are in 4th quadrant 2 3 4 3 sin cos 1 sin 2 1 5 5 5 2 17 15 8 15 sec cos sin 1 cos 2 1 15 17 17 17 3 15 4 8 45 32 13 Hence sin sin cos cos sin 5 17 85 85 5 17 20) tan x tan x tan x , as tan x tan x tan x tan x, as tan is negative in 2nd quadrant 3rd option cos 1 since 1 and domain of cos1 is 1,1 2 2 2 So cos 1 is not defined thus the equation has no solution 2 Last option 21)cos u cot u o 2nd option 2 cot u tan u so 5th option 2 sec u csc u so 3rd option 2 csc u sec u so 6th option 2 sin u cos u so 4th option 2 cos u sin u so 1st option 2 3)sin u v sin u cos v cos u sin v...5th option 2) tan cos u v cos u cos v sin u sin v....2nd option cos u v cos u cos v sin u sin v.....4th option sin u v sin u cos v cos u sin v....1st option tan u tan v ....3rd option 1 tan u tan v tan u tan v tan u v 6th option 1 tan u tan v tan u v a c 0.7 c sin sin sin 30 sin 90 0.7 0.7 1 c 1.4, as sin 90 1,sin 30 sin 30 1 / 2 2 5)given that 45, b 8.2, 90 Since the sum of angles of a triangle is 180 so 180 45 90 180 45 b a 8.2 a Then by sine rule we have a 8.2 sin sin sin 45 sin 45 6)given that a 29, c 96, 90 a c 29 96 29 By sine rule we have sin 17.5828 sin sin sin sin 90 96 Since the sum of angles of a triangle is 180 so 180 17.5828 90 180 72.4172 b c b 96 Then by sine rule we have b 96sin 72.4172 91.5150 sin sin sin 72.4172 sin 90 4)By sine rule in triangle ABC we have 7)given that 75.4, c 26, 90 Since the sum of angles of a triangle is 180 so 180 75.4 90 180 14.6 a c a 26 Then by sine rule we have a 6.5538 sin sin sin14.6 sin 90 8)Given Jack height is 6.5 ft. Distance between jack and pole is 60 ft and length is shadow is 24 ft. Let height of street light is x ft. In similar triangles ABC and ADF we have 84 6.5 BC DF 6.5 x x 22.75 23 ft AB AF 24 24 60 24 9)In triangle ADC we have angle of depression DAC 5 and height of ace is 3000 ft so DC 3000 ft So C 180 90 5 85 AD 3000 Then by sine rule in triangle ADC we have AD 34290.15691 ft sin85 sin 5 In triangle ABD we have the angle of elevation as BAD 15 So B 180 90 15 75 34290.15691 BD Then by sine rule in triangle ADB we have BD 9188.01985 ft sin 75 sin15 Thus total height of cliff is BD DC 9188.01985 3000 12188 ft 10)The person start at A and move in S25 E direction for 2 miles till point B. Then moves in N65 E direction for 6 miles till point C. So triangle ABC is right triangle as angle B 25 65 90 Hence AC2 22 62 40 AC 40 6.325 miles 11) x a cot for 0 , a 0 a a a a So , as a>0 and csc 2 1 cot 2 2 2 2 2 2 2 a x a a cot a 2 1 cot 2 a csc 1 sin so 2nd option is correct csc 1 1 x 116 45, as cos 45 2 2 12) 2 cos x 116 1 cos x 116 x 161 in interval 90,180 13)10sin 2 t 9sin t 7 0 sin t 9 81 280 9 19 1 7 , 2 10 20 2 5 Since 1 sin t 1 for all t so sin t 7 is not possible so discard this value 5 1 t 150 2 Hence sin t 14)sin 1 cos sin cos 1 1 2 sin 1 2 cos 1 2 sin cos 45 cos sin 45 1 2 1 45 45,135 0,90 2 Thus 2nd option is correct sin 45 2 2 ,cos 225 cos 180 45 cos 45 2 2 sin285 sin 60 225 sin 60 cos 225 sin 225 cos 60 15)Note that sin 225 sin 180 45 sin 45 2 1 6 2 so 1st option 2 2 4 16)csc0.24 csc 1.33079 sec1.33079 sec1.33 2 3 2 2 2 3 sin cos so 2nd option 8 8 2 8 18)cos21 sin 5 sin 21 cos5 cos 21 sin 5 sin 21 cos 5 , as cos x cos x 17)sin sin 5 21 , use sin a b sin a cos b cos a sin b sin16 hence 3rd option 19)Given , are in 4th quadrant 2 3 4 3 sin cos 1 sin 2 1 5 5 5 2 17 15 8 15 sec cos sin 1 cos 2 1 15 17 17 17 3 15 4 8 45 32 13 Hence sin sin cos cos sin 5 17 85 85 5 17 20) tan x tan x tan x , as tan x tan x tan x tan x, as tan is negative in 2nd quadrant 3rd option cos 1 since 1 and domain of cos1 is 1,1 2 2 2 So cos 1 is not defined thus the equation has no solution 2 Last option 21)cos u cot u o 2nd option 2 cot u tan u so 5th option 2 sec u csc u so 3rd option 2 csc u sec u so 6th option 2 sin u cos u so 4th option 2 cos u sin u so 1st option 2 3)sin u v sin u cos v cos u sin v...5th option 2) tan cos u v cos u cos v sin u sin v....2nd option cos u v cos u cos v sin u sin v.....4th option sin u v sin u cos v cos u sin v....1st option tan u tan v ....3rd option 1 tan u tan v tan u tan v tan u v 6th option 1 tan u tan v tan u v a c 0.7 c sin sin sin 30 sin 90 0.7 0.7 1 c 1.4, as sin 90 1,sin 30 sin 30 1 / 2 2 5)given that 45, b 8.2, 90 Since the sum of angles of a triangle is 180 so 180 45 90 180 45 b a 8.2 a Then by sine rule we have a 8.2 sin sin sin 45 sin 45 6)given that a 29, c 96, 90 a c 29 96 29 By sine rule we have sin 17.5828 sin sin sin sin 90 96 Since the sum of angles of a triangle is 180 so 180 17.5828 90 180 72.4172 b c b 96 Then by sine rule we have b 96sin 72.4172 91.5150 sin sin sin 72.4172 sin 90 4)By sine rule in triangle ABC we have 7)given that 75.4, c 26, 90 Since the sum of angles of a triangle is 180 so 180 75.4 90 180 14.6 a c a 26 Then by sine rule we have a 6.5538 sin sin sin14.6 sin 90 8)Given Jack height is 6.5 ft. Distance between jack and pole is 60 ft and length is shadow is 24 ft. Let height of street light is x ft. In similar triangles ABC and ADF we have 84 6.5 BC DF 6.5 x x 22.75 23 ft AB AF 24 24 60 24 9)In triangle ADC we have angle of depression DAC 5 and height of ace is 3000 ft so DC 3000 ft So C 180 90 5 85 AD 3000 Then by sine rule in triangle ADC we have AD 34290.15691 ft sin85 sin 5 In triangle ABD we have the angle of elevation as BAD 15 So B 180 90 15 75 34290.15691 BD Then by sine rule in triangle ADB we have BD 9188.01985 ft sin 75 sin15 Thus total height of cliff is BD DC 9188.01985 3000 12188 ft 10)The person start at A and move in S25 E direction for 2 miles till point B. Then moves in N65 E direction for 6 miles till point C. So triangle ABC is right triangle as angle B 25 65 90 Hence AC2 22 62 40 AC 40 6.325 miles 11) x a cot for 0 , a 0 a a a a So , as a>0 and csc 2 1 cot 2 2 2 2 2 2 2 a x a a cot a 2 1 cot 2 a csc 1 sin so 2nd option is correct csc 1 1 x 116 45, as cos 45 2 2 12) 2 cos x 116 1 cos x 116 x 161 in interval 90,180 13)10sin 2 t 9sin t 7 0 sin t 9 81 280 9 19 1 7 , 2 10 20 2 5 Since 1 sin t 1 for all t so sin t 7 is not possible so discard this value 5 1 t 150 2 Hence sin t 14)sin 1 cos sin cos 1 1 2 sin 1 2 cos 1 2 sin cos 45 cos sin 45 1 2 1 45 45,135 0,90 2 Thus 2nd option is correct sin 45 2 2 ,cos 225 cos 180 45 cos 45 2 2 sin285 sin 60 225 sin 60 cos 225 sin 225 cos 60 15)Note that sin 225 sin 180 45 sin 45 2 1 6 2 so 1st option 2 2 4 16)csc0.24 csc 1.33079 sec1.33079 sec1.33 2 3 2 2 2 3 sin cos so 2nd option 8 8 2 8 18)cos21 sin 5 sin 21 cos5 cos 21 sin 5 sin 21 cos 5 , as cos x cos x 17)sin sin 5 21 , use sin a b sin a cos b cos a sin b sin16 hence 3rd option 19)Given , are in 4th quadrant 2 3 4 3 sin cos 1 sin 2 1 5 5 5 2 17 15 8 15 sec cos sin 1 cos 2 1 15 17 17 17 3 15 4 8 45 32 13 Hence sin sin cos cos sin 5 17 85 85 5 17 20) tan x tan x tan x , as tan x tan x tan x tan x, as tan is negative in 2nd quadrant 3rd option cos 1 since 1 and domain of cos1 is 1,1 2 2 2 So cos 1 is not defined thus the equation has no solution 2 Last option 21)cos