Question

Base of right angled triangle is 5m and angle between base and hypotenuse is 60°. find the height of triangle​

Answer 1

$\bf{\underline{\underline{\bigstar\bigstar\: Solution : }}}$

$\:\:$

$\footnotesize{Cos\theta = \dfrac{Base}{hypotenuse}}$

$\footnotesize{\implies Cos{60}^{\circ} = \dfrac{5m}{Hypotenuse}}$

$\footnotesize{\implies \dfrac{1}{2} = \dfrac{5m}{Hypotenuse}}$

$\footnotesize{\implies Hypotenuse = \dfrac{5m}{\dfrac{1}{2}}}$

$\footnotesize{\implies Hypotenuse = \dfrac{5m \times 2}{1}}$

$\footnotesize{\implies Hypotenuse = \dfrac{10m}{1}}$

$\footnotesize{\implies Hypotenuse = 10m}$

$\:\:$

$\rule{200}3$

$\:\:$

$\underline{\bold{\bigstar\:Using\: Pythagoras\: theorem}}$

$\:\:$

$\footnotesize{\implies {Hypotenuse}^{2} = {Base}^{2} + {Height}^{2}}$

$\footnotesize{\implies {Hypotenuse}^{2} - {Base}^{2} = {Height}^{2}}$

$\footnotesize{\implies {(10m)}^{2} - {(5m)}^{2} = {Height}^{2}}$

$\footnotesize{\implies {100m}^{2} - {25m}^{2} = {Height}^{2}}$

$\footnotesize{\implies {75m}^{2} = {Height}^{2}}$

$\footnotesize{\implies \sqrt{{75m}^{2}} = Height}$

$\footnotesize{\implies 5\sqrt{3}\:m} = Height}$

$\footnotesize{\implies 8.66m = Height}$

Answer 2

$\large\bf\underline{Given:-}$

• Base of triangle = 5m
• Theta = 60°

$\large\bf\underline {To \: find:-}$

• height

$\huge\bf\underline{Solution:-}$

we know that,

• cot ∅ = base /perpendicular
• Base = 5m
• ∅ = 60°

• cot 60° = 1/√3

»»Cot 60° = 5m/height

»» 1/√3 = 5/height

»» height = 5√3

hence

• height of triangle is 5√3.

$\rule{200}3$

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}cot \theta&&\sqrt{3}&1 & \dfrac{1}{\sqrt{3}} & 0\end{tabular}}$