in the given fingers side BC of triangle ABC 6 standard to D if angle ACD is equal to 117 degree and angle abc is equal to 23° find Angle B A C and Angle A C B
Share
in the given fingers side BC of triangle ABC 6 standard to D if angle ACD is equal to 117 degree and angle abc is equal to 23° find Angle B A C and Angle A C B
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
Step-by-step explanation:
To solve this problem, we can use the properties of triangles. Let's denote the angles as follows:
- \(\angle ABC = A\)
- \(\angle BAC = B\)
- \(\angle ACB = C\)
From the information given:
1. \(BC\) is parallel to \(AD\) and \(BC\) is a transversal, so \(\angle ACD\) and \(\angle ABC\) are corresponding angles, and they are equal:
\(\angle ACD = \angle ABC = A = 117^\circ\)
2. \(\angle ABC = 23^\circ\)
Now, since the sum of angles in any triangle is \(180^\circ\), we can find angle \(C\) using the fact that:
\(A + B + C = 180^\circ\)
Substitute the known values:
\(117^\circ + 23^\circ + C = 180^\circ\)
Solve for \(C\):
\(140^\circ + C = 180^\circ\)
\(C = 180^\circ - 140^\circ\)
\(C = 40^\circ\)
Now that we know \(C\), we can find \(B\) using the fact that \(A + B + C = 180^\circ\):
\(117^\circ + B + 40^\circ = 180^\circ\)
\(B = 180^\circ - 117^\circ - 40^\circ\)
\(B = 23^\circ\)
Finally, we can find \(A\) using the fact that \(A + B + C = 180^\circ\):
\(A + 23^\circ + 40^\circ = 180^\circ\)
\(A = 180^\circ - 23^\circ - 40^\circ\)
\(A = 117^\circ\)
So, the angles of triangle ABC are:
- \(\angle ABC = A = 117^\circ\)
- \(\angle BAC = B = 23^\circ\)
- \(\angle ACB = C = 40^\circ\)