Advertisements
Advertisements
प्रश्न
Calculate x :
Advertisements
उत्तर
Let triangle be ABC and altitude be AD.
In ΔABD,
∠DBA = ∠DAB = 50° ...[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB ...[Exterior angle is equal to the sum of opp. interior angles]
∴ ∠CDA = 50° + 50°
⇒ ∠CDA = 100°
In ΔADC,
∠DAC = ∠DCA = x ...[Given AD = DC and angles opposite to equal sides are equal]
∴ ∠DAC + ∠DCA + ∠ADC = 180°
⇒ x + x + 100° = 180°
⇒ 2x = 80°
⇒ x = 40°
APPEARS IN
संबंधित प्रश्न
An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55°.
Find:
- ∠DCB
- ∠CBD
In the figure, given below, AB = AC.
Prove that: ∠BOC = ∠ACD.

Prove that a triangle ABC is isosceles, if: altitude AD bisects angles BAC.
Prove that a triangle ABC is isosceles, if: bisector of angle BAC is perpendicular to base BC.
In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.

In triangle ABC; AB = AC and ∠A : ∠B = 8 : 5; find angle A.
If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.
Use the given figure to prove that, AB = AC.
Prove that the medians corresponding to equal sides of an isosceles triangle are equal.
The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.
