Once a, B, C three good friends and took a taxi, told everyone to share the fare, a in the whole journey 1 Get off at 3 and get to 2 3 B also got off the bus, and finally C sat at the end of the line and paid the driver 90 yuan. Please calculate how much fare should be paid to C by a and B?

Once a, B, C three good friends and took a taxi, told everyone to share the fare, a in the whole journey 1 Get off at 3 and get to 2 3 B also got off the bus, and finally C sat at the end of the line and paid the driver 90 yuan. Please calculate how much fare should be paid to C by a and B?

According to the meaning of the title, the ratio of the distance traveled by a, B and C is 1:2:3, so the shared fare should be 1:2:3,
A to C: 90 × 1
1 + 2 + 3 = 15 yuan,
Payable by B to C: 90 × 2
1 + 2 + 3 = 30 yuan;
Answer: A should pay 15 yuan to C, B should pay 30 yuan to C

1 / 2 to 7 reduction ratio

1/2:7
=(1/2*2):(7*2)
=1:14

1.7; 1 / 2 reduction ratio

1.7 ：1/2
= 1.7:0.5
= 17:5
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The reduction ratio is 3 and 1 / 2:7

3 and 1 / 2: 7
=7/2：7
=1：2

Compare the size of sin (- 879), Tan (- 33 π / 8), cos (- 13 π / 10), and the detailed process

Using the period: sin (- 879) = sin (- 879 + 360 * 3) = sin201 = sin (180 + 21) = - sin21 ② Tan (- 33 π / 8) = Tan (- 33 π / 8 + 4 π) = Tan (- 1 / 8 π) ③ cos (- 13 π / 10) = cos (- 13 π / 10 + 2 π) = cos (7 / 10 π) = cos (π - 3 / 10 π) = - cos (54) = - cos (90-36)

How does tan2 compare to tan3?

1 = 57.3 degrees and so on, 2 = 114.6 degrees, 3 = 171.9 degrees
Because Tan function is an increasing function, tan3 is greater than tan2
The result is correct and the process is questionable. It is conditional that Tan function is an increasing function, which is the case in a certain domain

Mathematical calculation: TAN1 * tan2 * *tan89

The original formula = TAN1 * tan2 * *tan45*cot44*cot43*…… *cot1
=(tan1*cot1)*(tan2*cot2)*…… *tan45
=1

Confirmation: 3 + TAN1 ° · tan2 ° + tan2 ° · tan3 ° = tan3 ° tan1°．

It is proved that: 3 + TAN1 ° · tan2 ° + tan2 ° · tan3 °
=（1+tan1°•tan2°）+（1+tan2°•tan3°）+1
=tan2°−tan1°
tan(2−1)°+tan3°−tan2°
tan(3−2)°+1
=tan2°−tan1°+tan3°−tan2°
tan1°+1
=-1+tan3°
tan1°+1
=tan3°
tan1°
The original equation holds

Calculation: 1 / (1-tan / 8) - 1 / (1 + Tan / 8)

1 / (1-tan / 8) - 1 / (1 + Tan / 8) = 2tan / 8 / [1 - (Tan / 8) ^ 2] = Tan / 4 = 1

Tan school / 8-tan3 school / 8 =?

tan(π/8)-tan(3π/8)
=sin(π/8)/cos(π/8)-sin(3π/8)/cos(3π/8)
=[sin(π/8)cos(3π/8)-cos(π/8)sin(3π/8)]/[cos(π/8)cos(3π/8)]
=sin(π/8-3π/8)/{[cos(π/8+3π/8)+cos(π/8-3π/8)]/2}
=2sin(-π/4)/[cos(π/2)+cos(-π/4)]
=-2sin(π/4)/[0+cos(π/4)]
=-2•(√2/2)/(√2/2)
=-2
First, the chord is converted into a special angle by the sum and product formula of two angles. Of course, the half angle formula can also be used