In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD then DE : BC is
2 : 3
2 : 1
1 : 2
1 : 2
ABC is an isosceles triangle such that AB = AC and AD is the median to the base BC with ∠ABC = 35°.
Then ∠BAD is
35°
55°
70°
70°
A man goes 24 m due west and then 10 m due north. Then the distance of him from the starting point is
17 m
26 m
28 m
28 m
a and b are two sides of adjacent to the right angle of a right-angled and p is the perpendicular drawn to the hypotenuse from the opposite vertex. Then p2 is equal to
If the sum and difference of two angles are 135° and respectively, then the value of the angles in degree measure are
70°, 65°
75°, 60°
45°, 90°
45°, 90°
80°, 60°, 40°
70°, 50°, 60°
80°, 65°, 35°
80°, 65°, 35°
C.
80°, 65°, 35°
∠A + ∠B + ∠C = 180° ...(i)
(∠B - ∠C) - (∠A - ∠B ) = 30° - 15°
⟹ 2∠B - ∠A - ∠C = 15° ...(ii)
By adding (i) and (ii),
3∠B = 180° + 15° = 195°
⟹ ∠B = 65°
∠A - ∠B = 15°
⟹ ∠A = 15° + 65° = 80°
∠C = ∠B - 30°
= 65° - 30° = 35°
If ABC is an equilateral triangle and D is a point on BC such that then
AB : BD = 2 : 1
AB : BD = 1 : 2
AB : BD = 2 : 1
AB : BD = 2 : 1
In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e. hypotenuse. One of the acute angles must be
60°
30°
45°
45°