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In the figure, the sides $BC, CA$ and $AB$ of a $\triangle ABC$ have been produced to $D, E$ and $F$ respectively. If $\angle ACD = 105^o$ and $\angle EAF = 45^o$, find all the angles of the $\triangle ABC$."

Academic Mathematics NCERT Class 9

Given:

The sides $BC, CA$ and $AB$ of a $\triangle ABC$ have been produced to $D, E$ and $F$ respectively.

$\angle ACD = 105^o$ and $\angle EAF = 45^o$.

To do:

We have to find all the angles of the $\triangle ABC$.

Solution:

From the figure,

$\angle ACD + \angle ACB = 180^o$ (Linear pair)

$105° + \angle ACB = 180^o$

$\angle ACB = 180^o- 105^o = 75^o$

$\angle BAC = \angle EAF = 45^o$ (Vertically opposite angles)

$\angle BAC + \angle ABC + \angle ACB = 180^o$

$45^o + \angle ABC + 75^o = 180^o$

$120^o +\angle ABC = 180^o$

$\angle ABC = 180^o- 120^o$

$\angle ABC = 60^o$

Hence, $\angle ABC = 60^o, \angle BCA = 75^o$ and $\angle BAC = 45^o$.

Updated on: 10-Oct-2022

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