civonamzuk

Both posts are mine. Hopefully, one of them is correct...!? 

in a school, the ratio of the number of boys to the number of girls is 2:3 and the ratio of the number of girls to the number of teachers is 7:4. what is the ratio of the number of students to the number of teachers?

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B : G = 2 : 3                G : T = 7 : 4

2 : 3 = x : 7

x = 4 ²/₃

B : G : T = 4 ²/₃ : 7 : 4

(B + G) : T = 11 ²/₃ : 4  | *3

(B + G) : T = 35 : 12

In ΔJKL, l = 1.3 inches, j = 4.8 inches and ∠K=77°. Find ∠L, to the nearest 10th of a degree.

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Using the Law of cosines calculate the third side of a triangle JKL:

k² = l² + j² - 2lj*cos(K)           

And again, use the Law of cosines to find the measure of angle L:

l² = j² + k² - 2jk * cos(L)

Circle radius   R = 25         ( "R" is also the slant height of a cone ) 

Sector arc = 320º

Cone radius   r = 25(320/360)

Cone height    h = sqrt(R² - r²)

Cone volume    V = 1/3(πr²h)

Hello, catmg!

I like your method; good job!!! There're only a couple of errors:

Note that BAD * 2 ≠ BCD.                  ( ∠ABD * 2 = ∠CBD )

AC^2 = 5^2 + 6^2 - 2*5*6*cos(100.6719294 degrees)

AC = 8.491826135

Scale factor???

∠DAC + ∠AEF = 180º

∠DAB = ∠BAC 

∠BAC = (∠DAC + ∠AEF + 12) / 12

∠BAC = 16º

∠DAC = 32º

BL = x          CL = 12-x

We can solve this using the angle bisector theorem:

12/4 = (12-x)/x

12x = 4(12-x)

12x = 48-4x

16x = 48

x = 3                    BL = 3          CL = 9

x - (x/4 + 7) = 23

x = 40

Hunter had $40  

sinA = 2(cosA)

sin(A) / cos(A) = tan(A)

tan(A) = 2        ∠A = 63.43494882º

cos(A) = 0.447213595