Question

In Fig. 9.30, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC.

Answer 1

In Δ ABC $\angle ABC = 60\°$, $\angle BAC = 45\°$, $\angle ACB = 75\°$.

Step-by-step explanation:

Given:

$\angle ACD = 105\°$

$\angle EAF = 45\°$

We need to find all the angle of Δ ABC.

Solution:

Now we know that;

$\angle ACB + \angle ACD = 180\°$ ⇒ (Supplementary Angles)

Substituting the values we get;

$\angle ACB + 105\° = 180\°\\\\\angle ACB = 180\° -105\° = 75\°$

Now we know that Line CE and line BF intersect at point A

So we can say that;

$\angle EAF = \angle BAC$ ⇒ (Vertically opposite Angles)

$\angle EAF = \angle BAC = 45\°$

Now In Δ ABC,

"Sum of all angles of a triangle is 180°."

So we can say that;

$\angle ABC+\angle ACB+ \angle BAC = 180\°$

Substituting the values we get;

$\angle ABC+75\°+45\°=180\°\\\\\angle ABC +120\°=180\°\\\\\angle ABC = 180\°-120\°\\\\\angle ABC = 60\°$

Hence In Δ ABC $\angle ABC = 60\°$, $\angle BAC = 45\°$$\angle ACB = 75\°$.