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Step-by-step explanation:
in a right angle triangle ABC angle is equal to 90 degree and 5 sin square B + 7 cos square c + 4 upon 3 + 8 10 square 60 is equal to 7 upon 27 if AC is equal to 3 find the perimeter of triangle ABC
\frac{5Sin^2B + 7Cos^2C + 4}{3 + 8 Tan^260} = \frac{7}{27}
3+8Tan
2
60
5Sin
2
B+7Cos
2
C+4
=
27
7
Tan 60 = √3
Tan²60 = 3
Cos C = SinB as B+C = 90
=> \frac{5Sin^2B + 7Sin^2B + 4}{3 + (8\times 3)} = \frac{7}{27}
3+(8×3)
5Sin
2
B+7Sin
2
B+4
=
27
7
=> \frac{12Sin^2B + 4}{27} = \frac{7}{27}
27
12Sin
2
B+4
=
27
7
=> 12 Sin²B + 4 = 7
=> 12 Sin²B = 3
=> Sin²B = 3/12
=> Sin²B = 1/4
=> Sin B = 1/2
SinB = AC/BC
1/2 = 3/BC
=> BC = 6
AB² = BC² - AC²
=> AB² = 6² - 3²
=> AB² = 36 - 9
=> AB² = 27
=> AB = 3√3
=> AB = 5.2
Perimeter = AB + BC + AC
= 5.2 + 6 + 3
= 14.2
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