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

1 Answer

7 votes

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

User Tek Nath Acharya
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