Question

evaluate x and y from the given figure​

Answer 1

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

Consider,

In right-angle triangle ACD

AC = 10 m

CD = x m

∠ ADC = 60°

Thus,

$\rm \: cos60\degree = \dfrac{CD}{AC}$

$\rm \: \frac{1}{2} = \dfrac{x}{10}$

$\rm\implies \:x \: = \: 5 \: m$

Now, In right-angle BDC

$\rm \: siny\degree = \dfrac{DC}{BC}$

$\rm \: siny\degree = \dfrac{x}{5 \sqrt{2} }$

$\rm \: siny\degree = \dfrac{5}{5 \sqrt{2} }$

$\rm \: siny\degree = \dfrac{1}{ \sqrt{2} }$

$\rm \: siny\degree = sin45\degree$

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

$\rm\implies \:y = 45$

Hence,

$\rm\implies \:\boxed{\rm{  \:y = 45 \: }} \: \: and \: \: \boxed{\rm{  \:x \: = \: 5 \: \: }}$

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

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

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

Answer 2

In∆ACD

Finding 'x'

cos60°=(adj/hyp)

cos60°=x/10

1/2=x/10

x=5

y°=45°