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Paraphin [41]
3 years ago
6

How do I evaluate the six trigonomic functions of a right triangle with the hypotenuse of 25 and and the adjacent of 24

Mathematics
1 answer:
Korolek [52]3 years ago
7 0

First find the dimensions of the triangle by using the Pythagorean’s theorem.

a^2+b^2=c^2

*c is always the hypotenuse

24^2+b^2=25^2

576+b^2=625

625-576=b^2

49=b^2

Square root of 49= +or- 7b=+-7

*negative numbers are not often used in length for sides therefore we will be using the positive number.

Tan=opposite/adjacent= 7/24

Sin=opp/hypotenuse= 7/25

Cos = adjacent/hypo= 24/25

Cot=Tan^-1(inverse)= 24/7

Csc = Sin^-1= 25/7

Sec= Cos^-1= 25/24

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Please help with the 4th one
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You may want to have a look at the picture I attached.

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