
Trigonometry (11th Edition)
11th Edition
ISBN: 9780134217437
Author: Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher: PEARSON
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I need to know how I get 65 degrees out of sin(.909) because sin of (.909)=.79 and sin^-1=1.14 and I do not have arcsin on my calculator

Transcribed Image Text:9:38
AB =c = 12 ; AC =b; BC= a = 5
Applying pythagorous theorem in the given right
angle triangle
(Hypotenuse)? = (Base)2 + (Perpendicular)?
c2 = a² + b²
122 = 52 + b²
b2 = 144 -25 = 119
%3D
%3D
then b = V119
Step2
b)
According to the given figure
sinB = Perpendicular / Hypotenuse
= b/c
= V119/12
So,
sinB = V119/ 12
= 10.90/12 = 0.9090
%3D
B= sin'(0.9090) = 65.36º
Answer
%3D

Transcribed Image Text:10:13
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Expert Answer
Step 1
To find
arcsin(.9090)
Step 2
arcsin(.9090) = 65.36°
It is correct
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