Question

angle between base and hypotenuse is 60° and Base of right angled triangle is 8m. find the height of triangle.​

Answer 1

ɢɪᴠᴇɴ :-

★Base of triangle = 8m

★ Angle, ⌀ = 60°

ᴛᴏ ғɪɴᴅ:-

★ Height of the given triangle.

sᴏʟᴜᴛɪᴏɴ:-

⋆_______Some Information _________⋆

✰ tan ø = Perpendicular / Base

✰ ø =60°

✰ tan 60° =√3

✰ Base =8 m

______________⋆⋆

tan 60° = H/ 8

⟹ √3 = H/ 8

⟹ H = 8√3 m

Therefore,

height of the triangle is ☞ 8√3 m.

${\huge{\mathcal{\tt{Hope \ It \ Helps..!!!}}}}$

Answer 2

Given :

• Angle between base and hypotenuse is 60° and Base of right angled triangle is 8m.

To Find :

• The height of triangle = ?

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.9,0.7){\sf{\large{8 m}}}\put(7.4,2){\sf{\large{H}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{60^{\circ} }}\end{picture}$

Procedure:

$\dashrightarrow\:$$\sf tan \:\theta = \dfrac{Perpendicular}{Base}$

$\dashrightarrow\:$$\sf tan \:60^{\circ} = \dfrac{H}{8}$

$\dashrightarrow\:$$\sf \sqrt{3} = \dfrac{H}{8}$

$\dashrightarrow\:$$\sf H = 8\sqrt{3}$

$\dashrightarrow\:$$\sf H = 8 \times 1.73$

$\pink\dashrightarrow\:$$\pink{\sf H = 13.84\:m}$

Therefore, height of the triangle is 13.84 m.

Extra Brainly knowledge :

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$