Advertisements
Advertisements
प्रश्न
Find all the three angles of the ΔABC
Advertisements
उत्तर
∠A + ∠B = ∠ACD ...(An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x – 5 = 4x – 15
3x + 30 = 4x – 15
30 + 15 = 4x – 3x
45° = x
∠A = x + 35°
= 45° + 35°
= 80°
∠B = 2x – 5
= 2(45°) – 5°
= 90° – 5°
= 85°
∠ACD = 4x – 15
= 4(45°) – 15°
= 180° – 15°
= 165°
∠ACB = 180° – ∠ACD
= 180° – 165°
= 15°
∠A = 80°, ∠B = 85° and ∠C = 15°.
APPEARS IN
संबंधित प्रश्न
Fill in the blank to make the following statement true:
An exterior angle of a triangle is always ......... than either of the interior opposite angles.
In a Δ ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q, Prove that ∠BPC + ∠BQC = 180°.
In the given figure, AE bisects ∠CAD and ∠B= ∠C. Prove that AE || BC.
State exterior angle theorem.
In the given figure, side BC of ΔABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are
Find the unknown marked angles in the given figure:
Can a triangle together have the following angles?
33°, 74° and 73°
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
