Can someone help me with these questions please?
1 answer:
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<u>EXPLANATION</u><u>:</u>
In ∆ ABC , ∠ABC = 40°
- ∠CAB = x°
- & ∠ACD = 50°
∠ACD is an exterior angle formed by extending BC to D
We know that
The exterior angle of a triangle formed by extending one side is equal to the sum of the opposite interior angles.
∠ACD = ∠CAB + ∠ABC
⇛50° = x° + 40°
⇛x° = 50°-40°
<h3>⇛x° = 10°</h3>and
In ∆ ACD , AC = CD
⇛ ∠CDA = ∠CAD
Since the angles opposite to equal sides are equal.
Let ∠CDA = ∠CAD = A°
We know that
The sum of all angles in a triangle is 180°
In ∆ ACD,
∠CDA +∠CAD + ∠ACD = 180°
A°+A°+50° = 180°
⇛2A°+50° = 180°
⇛2A° = 180°-50°
⇛2A° = 130°
⇛A° = 130°/2
⇛A° = 65°
now,
∠CDA = ∠CAD = 65°
∠BAC + ∠CAD+y = 180°
Since angles in the same line
10°+65°+y = 180°
⇛75°+y =180°
⇛y = 180°-75°
<h3>⇛y = 105°</h3><u>Answer</u><u>:</u> Hence, the value of “x” & “y” will be 10° and 105° respectively.
