Advertisements
Advertisements
Question
In ΔABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
Options
35°
90°
70°
55°
Advertisements
Solution
In the given ΔABC, ∠B = ∠C . D is the ray extended from point A. AX bisects∠DAC and ∠DAX = 70°
Here, we need to find ∠ACB
As ray AX bisects ∠DAC
∠CAX = ∠DAX = 70°
Thus,
∠DAC = ∠DAX + ∠XAC
= 70° + 70°
= 140°
Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,
∠DAC = ∠B + ∠C
140° = 2∠C
`∠C = (140°)/2`
= 70°
Thus, ∠ACB = 70°
APPEARS IN
RELATED QUESTIONS
If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right triangle.
In the given figure, AB || DE. Find ∠ACD.
If the side BC of ΔABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.
Mark the correct alternative in each of the following:
If all the three angles of a triangle are equal, then each one of them is equal to
Find the unknown marked angles in the given figure:
Find x, if the angles of a triangle is:
2x°, 4x°, 6x°
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
S is any point on side QR of a ∆PQR. Show that: PQ + QR + RP > 2PS.
In the following figure, AB = BC and AD = BD = DC. The number of isosceles triangles in the figure is ______.
How is a vertex of triangle △ABC most accurately defined?
