if 16 cotA=12 find the value of
sin At Cost not
sin A-cos A. tou
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ce
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Answer:
Here we have our 13–14–15 triangle.
First of all, we must find cos(A)
Using the cosine law:
BC2=AB2+AC2−2∗AB∗AC∗cos(A)
142=132+152−2∗13∗15∗cos(A)
196=169+225−390∗cos(A)
198=390∗cos(A)
⇒cos(A)=198390
cos(A)=3365
Next, we’ll find sin(A)
Using the Pythagorean Identity:
sin2(A)+cos2(A)=1
⇒sin(A)=1−cos2(A)−−−−−−−−−√
sin(A)=1−(3365)2−−−−−−−−√
sin(A)=1−10894225−−−−−−−√
sin(A)=31364225−−−−√
sin(A)=5665
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