Question

please answer both the questions fast please ​

Answer 1

Answer:

option B

45 degree

Step-by-step explanation:

I hope it will help you

Answer 2

$\\ \large\bigstar \underbrace{ \textsf{ \textbf{ solution \: 1 : - }}}$

Observe that 2x and 2y are co-interior angles whose sum is equivalent to 180°

$\\ \quad \mapsto \sf 2x + 2y = 180 ^{0}$

$\\ \quad \mapsto \sf x + y = {90}^{0} - - - (1)$

In trangle ABC,

$\\ \quad \rightarrow \sf x + y + z = {180}^{0} \: \: \: \: \bigg \{ angle \: sum \: property\bigg \}$

$\\ \quad \rightarrow \sf {90}^{0} + z = {180}^{0} \: \: \: \: . \: . \: . \: from \: (1)$

$\\ \quad \rightarrow \sf z = {90}^{0}$

Now, z and 2a are in linear pair. Sum of the angles in a linear pair is equivalent to 180°

$\\ \quad \longrightarrow \sf z + 2a = {180}^{0}$

$\\ \quad \longrightarrow \sf {90}^{0} + 2a = {180}^{0}$

$\\ \quad\longrightarrow \sf 2a = {90}^{0}$

$\\ \quad \therefore \underline{ \boxed{ \bf a = {45}^{0} }} \bigstar$

hence, option (b) is correct

$\\ \large\bigstar \underbrace{ \textsf{ \textbf{ solution \: 2 : - }}}$

$\\ \quad \bullet \quad \sf x = 2 + \sqrt{5}$

$\\ \quad \mapsto\quad \sf x - \frac{1}{x} \\ \\ \quad \mapsto \quad \sf\frac{ {x}^{2} - 1 }{x}$

Now, substitute the values :-

$\\ \quad \mapsto \quad \sf\frac{ {(2 + \sqrt{5} )}^{2} - 1 }{2 + \sqrt{5} } \\ \\ \quad \mapsto \quad \sf\frac{4 + 5 + 4 \sqrt{5} - 1}{2 + \sqrt{5} } \\ \\ \quad \mapsto \quad \sf \frac{8 + 4 \sqrt{5} }{2 + \sqrt{5} } \\ \\ \quad \mapsto \quad \sf \frac{4 \cancel{(2 + \sqrt{5} )}}{ \cancel{2 + \sqrt{5} }} \\ \\ \quad \therefore \quad \boxed{ \boxed{ \bf 4} } \bigstar$

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