Determine the sin cos and tan of 3230118 to six places sin

. Determine the sin, cos and tan of 323°01\'18\" to six places. sin ~-ois 5. What is one-third of 193°1709\" 64 25\'43\" solutions. Report DMS angles to the nearest second, decimal degrees to six places, distances to the nearest 0.01 feet or nearest 0.001 mm. These units represent standard practice. If the last digit is a 5, round up if the next to the last digit is odd; and round down if the next to the last digit is even. What are all possible angles represented by tan +0.801010? Express your answers in DMS 6. 1.615572, cos--0.8 15128, and sin- 7. Using Fig. A, determine the length of (a) and (b) if (h) is 1000.00 feet and (A) is 35 8. Using Fig. A, determine the length of (a) if (h) is 1200.00 feet and (b) is 700.00 feet. 9. Using Fig. B, determine the length of (a). The length of (c) is 1000.00 feet and (C) 49-49\'59\" and (A) = 28°09\'29\". Using Fig. C, determine the length of (a). The length of(c) is 1500.00 feet and C 0 10. 1000200\" and A = 30°02\'00\". Figure A Figure B Figure C

Solution

9) we have given <A=28⁰09\'29\'\',<C=49⁰49\'59\'\' and length of c=1000ft

We know the formula sinA/a=sinB/b =sinC/c ---(1)

Converting angles A and C into degrees

<A= 28.15806⁰,<C=49.83306°

From 1 we get

sinA/a=sinC/c

sin(28.15806⁰)/a =sin(49.83306⁰)/1000 since c=1000ft

by doing cross multiplication

a=[1000*sin(28.15806⁰)]/sin(49.83306⁰) =1000*(0.47)/(0.76)=618.4ft

Therefore

The length of side a for given triangle is 618.4ft

13) we have given length c=1000ft and <A=37⁰,<B=87⁰

From given triangle is <A+<B+<C=180 implies <C=180-37-87=56⁰

By using formula sinA/a =sinB/b=sinC/c

sin(37)/a =sin(56)/1000 implies a=1000*(sin(37))/sin(56) =1000*(0.60)/(0.83) =722.9ft

a=722.9ft

sinB/b =sinC/c

sin(87)/b =sin(56)/1000 implies b=1000*sin(87)/sin(56)=1000*(0.99)/(0.83)=1192.7ft

Area of triangle is (1/2)*ab*sinC

(1/2)*ab*sinC =(1/2)*(722.9)*(1192.7)*sin(56)=(1/2)*(722.9)*(1192.7)(0.99)=426800 ft²

Therefore

the given tlriangle area is 426800 ft²