Question

construct ∆abc in which bc =5cm , angle CAB=120° and angle Abc=30° construt another triangle whose sides are 4/5 times thecrossponding side of ∆ABC

Answer 1

Take b=30° then take c=30° then measure u will get a=120°

Answer 2

$\triangle ABC$ is the required triangle and $\triangle A^{'}BC^{'}$ is another triangle.

Step-by-step explanation:

Given,

Construct $\triangle ABC$ where $BC=5\ cm$,$\angle CAB=120\ degree$ and $\angle ABC=30\ degree$

In triangle $\triangle ABC$,

$\angle CAB+\angle ABC+\angle ACB=180$

⇒$\angle ACB=(180-120-30)$

$\angle ACB=30\ degree$

Following steps:

• Draw a line segments $BC=5\ cm$ using scale.

• At $B$, draw an angle $30\ degree$ like $\angle CBX=30\ degree$

• Also at $C$, draw an angle $30\ degree$ like $\angle BCY=30\ degree$  
• Line $BX$ and $CY$ meet at $A$

Now, $\triangle ABC$ is the required triangle.

• Line $BC$ divide $5$ equal parts and every parts is $(\frac{5}{5})=1\ cm$ using scale and naming last point as $C^{'}$

• Draw parallel line with $AC$ through point $C^{'}$ which meet at $A^{'}$ on $AB$  

Now, $\triangle A^{'}BC^{'}$ is another triangle.