Question

find the value of angle A
​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that the given right-angled triangle is XYZ, right angled at X, such that XY = y and XZ = x and ∠Y = A and ∠X = 90°.

Now, In right triangle XYZ

$\rm \: tanA \: = \: \dfrac{XZ}{XY}$

$\rm \: tanA \: = \: \dfrac{x}{y} - - - (1)$

Now, Solution (i), when x = y

$\rm \: tanA \: = \: \dfrac{y}{y}$

$\rm \: tanA \: = \: 1$

$\rm \: tanA \: = \: tan45 \degree$

$\rm\implies \:A \: = \: 45 \degree$

Now, Solution (ii)

$\rm \: When \: x = \sqrt{3} \: y$

So, equation (1), can be rewritten as

$\rm \: tanA \: = \: \dfrac{ \sqrt{3} \: y}{y}$

$\rm \: tanA \: = \: \sqrt{3}$

$\rm \: tanA \: = \: tan60 \degree$

$\rm\implies \: A \: = \: 60 \degree$

Now, Solution - (iii)

$\rm \: When \: y = \sqrt{3} \: x$

So, equation (1) can be rewritten as

$\rm \: tanA \: = \: \dfrac{x}{ \sqrt{3} \: x}$

$\rm \: tanA \: = \: \dfrac{1}{ \sqrt{3}}$

$\rm \: tanA \: = \: tan30 \degree$

$\rm\implies \: A \: = \: 30 \degree$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

$\boxed{\underline{\underline{\text{Solution:}}}}$

we have,

$\\ \tt(i) x=y^{\circ}$

$\\ \\ \tt\tan A=\frac{x}{y}=\frac{x}{x}=1$

$\\ \\ \tt\tan A=\tan 45^{\circ}$

$\\ \\ \color{purple}\tt\angle A=45^{\circ}$

$\\ \\ \rule{200pt}{2pt}$

$\\ \\ \begin{array}{l} \sf(ii) \: \: \: \tt x=\sqrt{3} y \\ \\ \\ \tt \tan A=\dfrac{x}{y}=\dfrac{\sqrt{3} y}{y}=\sqrt{3} \\ \\ \\ \tt \tan A=\tan 60^{\circ} \\ \\ \\ \red{ \tt \angle A = 60 {}^{ \circ} } \end{array}$

$\\ \\ \rule{200pt}{2pt}$

$\begin{array}{l} \tt(ii) \sqrt{3} x=60^{\circ} \\ \\ \\ \tt\tan A=\dfrac{x}{y}=\dfrac{x}{\sqrt{3} x}=\dfrac{1}{\sqrt{3}} \\ \\ \\ \tt \tan A=\tan 30^{\circ} \\ \\ \\ \color{darkcyan}\tt \angle A=30^{\circ} \end{array}$

Hence, it is required solution.