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CAN SOMEONE HELP WITH THIS QUESTION?✨

CAN SOMEONE HELP WITH THIS QUESTION?✨-example-1
User Arnon Axelrod
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1 Answer

21 votes
21 votes

Explanation:

as this is a right-angled triangle, we use Pythagoras to get also c :

c² = a² + b² = 2² + 7² = 4 + 49 = 53

c = sqrt(53)

we know, sine = opposite/Hypotenuse.

so,

sin(A) = 2/sqrt(53) = 0.274721128...

from the norm circle we know cosine is the other leg of the right-angled triangle :

cos(A) = 7/sqrt(53) = 0.961523948...

tan(A) = sin(A)/cos(A) = 2/7 = 0.285714286...

sec(A) = 1/cos(A) = sqrt(53)/7 = 1.040015698...

csc(A) = 1/sin(A) = sqrt(53)/2 = 3.640054945...

cot(A) = 1/tan(A) = cos(A)/sin(A) = 7/2 = 3.50

oh, and FYI :

A = 15.9453959...°

User Chaseadamsio
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2.7k points