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
Two angles of a triangle are equal and the third angle is greater than each of those angles
by 30°. Determine all the angles of the triangle.
Can a triangle have All angles less than 60° Justify your answer in case.
Compute the value of x in the following figure:

In Δ ABC, BD⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° − A.
In the following, find the marked unknown angle:

Find, giving a reason, the unknown marked angles, in a triangle drawn below:

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.
7 cm, 24 cm, 25 cm
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
The correct statement out of the following is 
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
