if sec theta = 3/2 find all trigonometric ratios
Answers
Answer:
given: sec theta = √3/2
sec theta= Base/Perpendicular
therefore, base =√3 and perpendicular=2
then by PYTHAGORAS THEOREM
(H)²=(B)²+(P)²
then H =√7
sin theta= P/H = 2/√7
cos theta= B/H = √3/√7
tan theta= P/B = 2/√3
cosec theta= H/P = √7/2
sec theta= H/B =√7/√3
Answer:
Explanation:
Given sec theta = 3/2
Consider a triangle ABC with <B=90⁰ and <C = theta
Sec theta = 3/2 = AC/BC
Let AC be 3k and BC be 2k
By pythagoras theorem, AB² + BC² = AC²
AB²= AC²-BC²=(3k)²-(2k)²
= 9k²-4k²=5k²
AB = root5 k
Sin thetha = opposite/ hypotenuse = root 5 k /3k = root 5 /3
Cos theta = 2k/3k = 2/3
Tan theta = root 5 k / 2k = root5/2
Cosec theta = 1/sin theta = 3/root 5
Cot theta = 1/tan theta = 2/root 5
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