Two angles of a triangle are equal and the third angle measures 70°. Find the measure of each of the unknown angles.
Let the measure of each unknown angle be x°. We know that the sum of the angles of a triangle is 180°.
∴ x + x + 70 = 180°
⇒ 2x = (180° – 70°)
⇒ 2x = 110°
⇒ x = 55°
Hence each unknown angle is 55°.
In a ∆XYZ, if ∠X = 90° and ∠Z = 48°, find ∠Y.
We know that the sum of the angles of a triangle is 180°.
Each of the two equal angles of an isosceles triangle is twice the third angle. Find the angles of the triangle.
Let the third angle be x°.
We know that the sum of the angles of a triangle is 180°.
∴ 2x + 2x + x = 180°
⇒ 5x = 180°
⇒ x = 36°
So, the angles are (2 × 36)°, (2 × 36)° and (x)°
Hence, the angles of the triangles are 72°, 72°, 36°.
What is the measure of each angle of an equilateral triangle?
All the angles of equilateral triangle are equal.
Find the angles of a triangle which are in the ratio 4 : 3 : 2.
Let the measure of the given angles of the triangle be (4x)°, (3x)° and (2x)° respectively.
4x + 3x + 2x = 180°
⇒ 9x = 180°
So, the angles measure (4×20)°, (3 × 20)° and (2 × 20)°
Hence, the angles of the triangle are 80°, 60°, 40°.
In the given figure, find the values of x and y.
We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
In the given figure, find the values of x and y.
We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.
The sum of all the angles of a triangle is 180°.
x + y + 68 = 180° ⇒ 62° + y + 68° = 180°
⇒ 130° + y = 180°
y = 180° – 130°
= 50°
In the figure given alongside, x : y = 2 : 3 and ∠ACD = 130° find the values of x, y and z.
Let x = 2t, y = 3t, then 2t + 3t = 130° ⇒ t = 26°
∴ Sum of all angles of a triangle is 180°
A man goes 24m due east and then 10m due north. How far is he away from his initial position?
Let O be the initial position of the man. Let he cover OA = 24m due east and then AB = 10m due north.
Finally, he reaches the point B. Join OB.
OB^{2 }= (OA^{2} + OB^{2}) = {(24)^{2} + (10)^{2}} m^{2}
= (576 + 100) m^{2} = 676 m^{2}
Hence, the man is at a distance at 26m from his initial position.
The lengths of the sides of two triangles are given below. Which of them is right – angled?
(i) a = 8 cm, b = 5 cm, and c = 10 cm
(ii) a = 7 cm, b = 24 cm, and c = 25 cm
(i)
Here a = 8 cm, b = 5cm and c = 10c,
The largest side is c = 10cm.
a^{2} + b^{2} = {(8)^{2} + (5)^{2}} cm^{2}
= m (64 + 25) cm^{2} = 89 cm^{2} ≠ (10)^{2 }cm^{2}
a^{2} + b^{2} ≠ c^{2}
∴ Given triangle is not right angled.
(ii) Here a = 7 cm, b = 24 cm and c = 25 cm
The largest side is c = 25 cm
a^{2} + b^{2} = {(7)^{2} + (24)^{2}} cm^{2} = (49 + 576) cm^{2}
= 625 cm^{2} = (25 cm)^{2} = c^{2}
⇒ a^{2} + b^{2} = c^{2}
Given triangle is right angled.
Two poles of height 9 cm and 14m stand upright on a plane ground. If the distance between their feet is 12 m. Find the distance between their tops.
Let AB and CD be the given poles such that
AB = 9 m, CD = 14 m and AC = 12 m. Join BD.
From B, draw BL ⊥ CD.
DL = (CD – CL) = (CD – AB)
= (14 – 9) m = 5 m
BL = AC = 12 m
Now, in right ∆BLD, by Pythagoras theorem.
We have
BD^{2} = BL^{2} + DL^{2} = {(12)^{2} + (5)^{2}}m^{2}
= (144 + 25) m^{2} = 169 m^{2}
Two circles are congruent if they have
Two lines segments are congruent if they have
In a ∆ABC it is given that ∠B = 37° and ∠C = 29°. Then ∠A = ?
We know that the sum of the angles of a triangle is 180°.
In a ∆ABC, if 2∠A = 3, ∠B = 6∠C then ∠B = ?
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