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4 years ago

1) - 2 - 5 = 2) 10 - (-2) = 3) -25 - 5 = 4) How are the three problems above different from this problem: -12 + 5

1 answer:

5 0

They’re all different because questions 1-3 is like adding. While question 4, you’re answering it as if you’re subtracting. So -12+5=-7.

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Part 1) $sin(B)=\frac{21}{29}$

Part 2) $csc(A)=\frac{29}{20}$

Part 3) $cot(A)=\frac{21}{20}$

Step-by-step explanation:

The complete question is

Consider this right triangle. 21 29 20 Write the ratio equivalent to: Sin B - CscA- Cot B

The picture of the question in the attached figure

Part 1) Write the ratio equivalent to: Sin B

we know that

In the right triangle ABC

$sin(B)=\frac{AC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the values

$sin(B)=\frac{21}{29}$

Part 2) Write the ratio equivalent to: Csc A

we know that

In the right triangle ABC

$csc(A)=\frac{1}{sin(A)}$

$sin(A)=\frac{BC}{AB}$ -----> by SOH (opposite side divided by the hypotenuse)

substitute the values

$sin(A)=\frac{20}{29}$

therefore

$csc(A)=\frac{29}{20}$

Part 3) Write the ratio equivalent to: Cot A

we know that

In the right triangle ABC

$cot(A)=\frac{1}{tan(A)}$

$tan(A)=\frac{BC}{AC}$ -----> by TOA (opposite side divided by the adjacent side)

substitute the values

$tan(A)=\frac{20}{21}$

therefore

$cot(A)=\frac{21}{20}$

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3 years ago