Advertisements
Advertisements
प्रश्न
Find, giving a reason, the unknown marked angles, in a triangle drawn below:
Advertisements
उत्तर
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
∴ 110° = x + 30° (by property)
⇒ x = 110°- 30°
x = 80°
APPEARS IN
संबंधित प्रश्न
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 10.26). Show that the line PQ is perpendicular bisector of AB.
In the given figure, AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.
In ΔABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
In ΔPQR, If ∠R > ∠Q then ______.
In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
In ∆ABC, C = 56° C = 56° ∠B = ∠C and ∠A = 100° ; find ∠B.
Find, giving a reason, the unknown marked angles, in a triangle drawn below:
Which two triangles have ∠B in common?
The sides of a triangle must always be connected so that:
