Advertisements
Advertisements
प्रश्न
In the following, find the marked unknown angle:

Advertisements
उत्तर
Since, sum of all angles of triangle = 180°
Hence, 60° + 45° + 20° + x = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125°
⇒ x = 55°
APPEARS IN
संबंधित प्रश्न
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, AB || DE. Find ∠ACD.

Is the following statement true and false :
All the angles of a triangle can be less than 60°
In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

In the given figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

One angle of a triangle is 61° and the other two angles are in the ratio `1 1/2: 1 1/3`. Find these angles.
Find x, if the angles of a triangle is:
2x°, 4x°, 6x°
The length of the three segments is given for constructing a triangle. Say whether a triangle with these sides can be drawn. Give the reason for your answer.
8.4 cm, 16.4 cm, 4.9 cm
P is a point on the bisector of ∠ABC. If the line through P, parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆OCD is an isosceles triangle.
