Advertisements
Advertisements
प्रश्न
An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to
पर्याय
75°
80°
80°
40°
50°
Advertisements
उत्तर
n the ΔABC, CD is the ray extended from the vertex C of ΔABC. It is given that the exterior angle of the triangle is 100° and two of the interior opposite angles are equal.
So, ∠ACD = 100° and A = ∠B
So, now using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get.
In ΔABC
∠A + ∠B = ∠ACD
∠2A = 100°
`∠A = (100°)/2`
∠A = 50°
∠A = ∠B = 50°
Therefore, each of the two opposite interior angles is 50°.
APPEARS IN
संबंधित प्रश्न
ABC is a triangle in which ∠A — 72°, the internal bisectors of angles B and C meet in O.
Find the magnitude of ∠BOC.
In a ΔABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
In a Δ ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
If two acute angles of a right triangle are equal, then each acute is equal to
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = \[\frac{1}{2}\]∠A is equal to
Find the value of the angle in the given figure:
Classify the following triangle according to angle:
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
What is common in the following figure?
 |
 |
| (i) | (ii) |
Is figure (i) that of triangle? if not, why?
Can we have two acute angles whose sum is a right angle? Why or why not?
