А 15 m
12 m
B
a.
c.
sin A=
3
tan A =
59
4
4
5
sec A =
sin A=
3
tan A =
4'
sec A=
4
3
4
5
4
5
4
cos A =
3
cos A =
5, cot A=
3s csc A=
5, cot A =
32PCSC A =
b.
4
4
5
d.
sin A = 5, tan A = 3, sec A =
5
4
4
sin A =
31
tan A = 3, sec A =
3
3
3
3
3
cOS A =
5'
cot A =
4
CSC A =
4
5
COS A =
4
cot A =
4'
CSC A =

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

answer: x=60