Question

Q16 In any triangle ABC,
Sin(A+B)​

Answer 1

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