Question

please answer this 3 question with (step by step)please please please please please please please please please please please please please please please please​

Answer 1

Answer:

1)

Given:

tan 53° . tan 50° . tan 37° . tan 40° = 1

We know that tan and cot are complementary (i.e. sum is equal to 90°)

$\tan(90 ^{ \circ} - \theta) = \cot \: \theta$

$\therefore \: \tan( {90}^{\circ} - {37}^{\circ} ) = \cot \: {37}^{\circ}$

and

$\tan \: ( {90}^{\circ} - {50}^{\circ} ) = \cot \: {50}^{\circ}$

Therefore,

LHS

$\tan \: {37}^{\circ} \times \cot \: {37}^{\circ} \times\tan \: {50}^{\circ} \times \cot \: {50}^{\circ}$

We know that tan and cot are reciprocal of each other.

$\tan \: {37}^{\circ} \times \frac{1}{\tan {37}^{\circ} } \times \tan {50}^{\circ} \times \frac{1}{\tan {50}^{\circ} }$

$1 \times 1$

$1$

Therefore,

LHS = RHS

2)

Given:

Two points :

(-15,0) and (6,0)

Here,

$x_1 = -15$

$y_1 = 0$

$x_2 =6$

$y_2 = 0$

$\tt \: Distance \: Formula = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$

Substituting the values , we get :

$\tt \: distance = \sqrt{(6 - ( - 15))^{2} + {(0 - 0)}^{2} }$

$\tt \: distance \: = \sqrt{ {21}^{2} }$

$\tt\therefore \: distance = 21 \: units$

3)

Given:

$\frac{4 {x}^{2} + 3 \sqrt{3}x + 7 }{7} = 0$

Cross multiplying, we get:

$4 {x}^{2} + 3 \sqrt{3} x + 7 = 0$

Here,

a = 4

b = 3√3

c = 7