Question

If the hypotenuse of a 30°-60°-90° Triangle is 10√2, find the length of the other two sides.

Answer 1

Let us consider a 30°-60°-90°△ABC right angled at B in which ∠C = 30⁰ and ∠A = 60⁰ with hypotenuse AC = 10√2 units.

Solution:-

In △ABC,

`\longrightarrow`sin ∠C =
`\sf (Perpendicular)/(Hypotenuse)`

`\longrightarrow`sin ∠C =
`\sf (AB)/(AC)`

`\longrightarrow`sin 30⁰ =
`\sf (AB)/(10√(2))`

`\longrightarrow`
`\sf (1)/(2)= (AB)/(10√(2))`

`\longrightarrow`
`\sf AB = (10√(2))/(2)`

`\longrightarrow`
`\sf AB = 5√(2)\:units`

`\\`

In △ABC,

`\longrightarrow`cos ∠C =
`\sf (Base)/(Hypotenuse)`

`\longrightarrow`cos ∠C =
`\sf (BC)/(AC)`

`\longrightarrow`cos 30⁰ =
`\sf (BC)/(10√(2))`

`\longrightarrow`
`\sf (√(3))/(2)= (BC)/(10√(2))`

`\longrightarrow`
`\sf BC = (10√(2)* √(3))/(2)`

`\longrightarrow`
`\sf BC = (10√(2* 3))/(2)`

`\longrightarrow`
`\sf BC = (10√(6))/(2)`

`\longrightarrow`
`\sf BC = 5√(6)\: units`

Thus , the length of other two sides of triangles are 5√2 and 5√6 units.