Question

if cot0 = 5/12 the find 5sin0-3cos0/5sin0+3cos0​

Answer 1

Step-by-step explanation:

cotθ = 5/12

tanθ = 12/5

h = 13

$\frac{3cos θ}{5sinθ+3cosθ} \\\frac{3(\frac{5}{13} )}{5(\frac{12}{13} )+3(\frac{5}{13})} \\\frac{\frac{15}{13} }{\frac{60}{13}+\frac{15}{13} }\\\frac{\frac{15}{13} }{\frac{75}{13} }\\\frac{15}{75} =\frac{1}{5}$

Answer 2

Answer:

cotΘ = 5/12

5/12 = adjacent/ opposite

In right triangle ABC,

AB² + BC² = AC²

12² + 5² = AC²

144 + 25 = AC²

AC² = 169

AC = 13 units

sinΘ = 12/13, cosΘ = 5/13

5sinΘ = 5(12)/ 13 = 60/13, 3cosΘ = 3(5)/13 = 15/13

5sinΘ - 3cosΘ = 60-15/13 = 45/13   ------- 1

5sinΘ + 3cosΘ = 60+15/13 = 75/13  -------- 2

1 ÷ 2

5sinΘ - 3cosΘ/5sinΘ + 3cosΘ = 45/13 * 13/75

= 45/75

= 3/5.

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