Question

Can someone help me answer this question? I’m so confused.

Answer 1

Explanation:

Required Answer:-

Let the shortest leg=x

other leg=2x

• Hypontenuse=h=5cm

According to Pythagoras thereon

`{\boxed{\sf p^2+b^2=h^2}}`

• Substitute the values

`{:}\dashrightarrow`
`\sf (x)^2+(2x)^2=(5)^2`

`{:}\dashrightarrow`
`\sf x^2+2x^2=25`

`{:}\dashrightarrow`
`\sf 3x^2=25`

`{:}\dashrightarrow`
`\sf x^2={(25)/(3)}`

`{:}\dashrightarrow`
`\sf x^2=8.3`

`{:}\dashrightarrow`
`\sf x=\sqrt {8.3}`

`{:}\dashrightarrow`
`\sf x=2.8cm`

`{:}\dashrightarrow`
`\sf x=3cm (Approx)`

`\therefore`The shorter leg is 3cm long

Learn more:-

Trigonometric Values table:-

`\Large {\begin{tabular} \cline{1-6} \theta & \sf 0^(\circ) & \sf 30^(\circ) & \sf 45^(\circ) & \sf 65^(\circ) & \sf 90^(\circ) \\ \cline{1-6} $ \sin $ & 0 & $(1)/(2 )$ & $(1)/( √(2) )$ & $( √(3))/(2)$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ ( √( 3 ))/(2) } $ & $ (1)/( √(2) ) $ & $ ( 1 )/( 2 ) $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ (1)/( √(3) ) $ & 1 & $ √(3) $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ √(3) $ & 1 & $ (1)/( √(3) ) $ &0 \\ \cline{1 - 6} \sec & 1 & $ (2)/( √(3)) $ & $ √(2) $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ √(2 ) $ & $ ( 2 )/( √( 3 ) ) $ & 1 \\ \cline{1 - 6}\end{tabular}}`

Answer 2

the shorter leg is 2 cms long

Explanation: