Advertisements
Advertisements
Question
The angle of a vertex of an isosceles triangle is 100°. Find its base angles.
Advertisements
Solution
In Δ ABC,
∵ AB = AC.
∴ ∠B = ∠C
But ∠A = 100°
and ∠A + ∠B + ∠C = 180° ..........(Angles of a triangle)
⇒ 100° + ∠B + ∠B = 180°
⇒ 2 ∠B = 180°− 100°
⇒ 2 ∠B = 80°
∴ ∠B =`(80°)/2=40°`
Hence ∠B = ∠C = 40°
APPEARS IN
RELATED QUESTIONS
In Fig. 10.25, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
In the given figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.
Define a triangle.
In the given figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.
The exterior angle of a triangle is equal to the sum of two
Q is a point on the side SR of a ∆PSR such that PQ = PR. Prove that PS > PQ.
The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in the following figure. Prove that the image is as far behind the mirror as the object is in front of the mirror.
[Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection].
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 straight angle? Why or why not?
What is the sum of the measures of the interior angles of any triangle, △ABC?
