Section A 1 Mark Each
1. The bisectors of acute angles B and C of a right ABC meet at P. Then BPC =
(A) 135 (B) 90 (C) 45 (D) 20
2. The angle which is 8 times its complement is
(A) 80 (B) 72 (C) 90 (D) 160
3. An exterior angle of a triangle is 100 and its two interior opposite angles are equal.
Each of these equal angles is
(A)
1
37
2
 (B)
1
52
2
 (C)
1
72
2
 (D) 75
4. If 3 sides of a triangle are produced in order then the sum of three exterior angles =
(A) 180 (B) 720 (C) 90 (D) 360
5. The sum of the acute angles of an obtuse triangle is 80 and their difference is10 .
The lsmallest angles is
(A) 35 (B) 45 (C) 50
(D) 25
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2
Section B 2 Marks Each
7. Find value of ‘a’ for which l ll m.
Section C 3 Marks Each
8. In the given figure, AB and BC are two plane mirrors
perpendicular to each other. Show that the incident ray
PQ is parallel to reflected ray RS.
9. In triangle ABC bisectors of exterior angles B and C
Intersect each other at O. Prove
A
BOC 90
2
 
  
10. Arms of one angle are respectively parallel to arms of
other angle . Prove angle are equal or supplementary.
Section D 4 Marks Each
11. Prove midpoint of hypotenuse of a right triangle is equidistant from the three
vertices.

2)

Let the complement angle be x 

According to the given condition,

angle 8 times its complement = 8x

⇒x + 8x = 90°

⇒ 9x = 90°

⇒ x = 10°

So, the angle 8 times its complement = 8x = 8 × 10° = 80°

 

5)

Let the smaller acute angles be y and larger be x.

According to the given condition,

x + y =  80°  ... (1)

and x - y = 10°  ... (2)

On adding (1) and (2), we get

2x = 90°

⇒ x = 45°

On putting x in (1),we get

y = 80° - 45° = 35°

Hence, the smallest angle is of 35°.

 

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