The sides of a triangle are $$\sin \alpha ,\,\cos \alpha $$ and $$\sqrt {1 + \sin \alpha \cos \alpha } $$ for some $$0 < \alpha < {\pi \over 2}$$. Then the greatest angle of the triangle is
A
$${150^ \circ }$$
B
$${90^ \circ }$$
C
$${120^ \circ }$$
D
$${60^ \circ }$$
Explanation
Let $$a = \sin \alpha ,b = \cos \alpha $$
and $$c = \sqrt {1 + \sin \alpha \cos \alpha } $$
Clearly $$a$$ and $$b < 1$$ but $$c > 1$$
as $$\,\,\,\sin \alpha > 0$$ and $$\cos \alpha > 0$$
$$\therefore$$ $$c$$ is the greatest side and greatest angle is $$C$$
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \circ }$$. The breadth of the river is
A
$$60\,\,m$$
B
$$30\,\,m$$
C
$$40\,\,m$$
D
$$20\,\,m$$
Explanation
From the figure
$$\tan {60^ \circ } = {y \over x}$$
$$ \Rightarrow y = \sqrt {3x} .......\left( 1 \right)$$
$$\sqrt 3 x = {{x + 40} \over {\sqrt 3 }} \Rightarrow x = 20m$$
3
AIEEE 2003
MCQ (Single Correct Answer)
If in a $$\Delta ABC$$ $$a{\cos ^2}\left( {{C \over 2}} \right) + {\cos ^2}\left( {{A \over 2}} \right) = {{3b} \over 2},$$ then the sides $$a, b$$ and $$c$$
In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,
$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is