Question

The value of (4-〖sin〗^2 45°)/(cotk tan60°) is 3.5 , What is the value of k? *​

Answer 1

Answer

x = 60°

process :-

see the attachment

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ \frac{4 - { \sin}^{2} {45}^{ \circ} }{ \cot k \tan {60}^{ \circ} } = 3.5}$

TO DETERMINE

The value of k

EVALUATION

$\displaystyle \sf{ \frac{4 - { \sin}^{2} {45}^{ \circ} }{ \cot k \tan {60}^{ \circ} } = 3.5}$

$\displaystyle \sf{ \implies \frac{4 - { \big( \frac{1}{ \sqrt{2} } \big)}^{2} }{ \cot k \times \sqrt{3} } = 3.5}$

$\displaystyle \sf{ \implies \frac{4 - \frac{1}{2} }{ \cot k \times \sqrt{3} } = 3.5}$

$\displaystyle \sf{ \implies \frac{ \frac{8 - 1}{2} }{ \cot k \times \sqrt{3} } = 3.5}$

$\displaystyle \sf{ \implies \frac{ \frac{7}{2} }{ \cot k \times \sqrt{3} } = 3.5}$

$\displaystyle \sf{ \implies \frac{ 3.5 }{ \cot k \times \sqrt{3} } = 3.5}$

$\displaystyle \sf{ \implies \cot k \times \sqrt{3} } =1$

$\displaystyle \sf{ \implies \cot k } = \frac{1}{ \sqrt{3} }$

$\displaystyle \sf{ \implies \cot k } = \cot {60}^{ \circ}$

$\displaystyle \sf{ \implies k = {60}^{ \circ} }$

FINAL ANSWER

Hence the required value of k = 60°

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