In a ∆ABC, ∠A – ∠B = 33° and ∠B – ∠C = 18°, then ∠B = ?
∠A = 33° + ∠B and ∠C = ∠B – 18°
∴ (33° + ∠B) + ∠B + (∠B – 18°) = 180°
⇒ 3∠B = 165° 17. (d) ⇒ ∠B = 55°
The sum of all angles of a triangle is
The sum of all the angles of a triangle is 180°.
In the given figure, what value of x will make AOB a straight line?
In the given figure, AOB is a straight line,
∠AOC = (3x – 8)°, ∠COD = 50° and ∠BOD = (x + 10)°. The value of x is
Here, (3x – 8)° + 50° + (x + 10)° = 180°
⇒ 4x + 52 = 180°
⇒ 4x = 180 – 52
⇒
⇒ x = 32
In ∆ABC, ∠B = 90°, AB = 5cm and AC = 13cm then BC = ?
By Pythagoras theorem,
AB^{2} + BC^{2} = AC^{2} ⇒ 5^{2} + BC^{2} = 132
= 12 cm
∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm then AB = ?
The angles of a triangle are (3x)°, (2x – 7)° and (4x – 11)°. Then x = ?
According to question
3x + (2x – 7) + (4x – 11) = 180°
⇒ 9x – 18 = 180
⇒ 9x = 198
⇒ x = 22
The supplement of 45° is
An angle is one – fifth of its supplement. The measure of the angle is
In the given figure, AOB is a straight line, ∠AOC = 68° and ∠BOC = x°, the value of x is.
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