41.
In a $$\vartriangle ABC,$$ $$I$$ is the incentre. The ratio $$IA : IB : IC$$ is equal to
A.
$${\text{cosec}}\frac{A}{2}:{\text{cosec}}\frac{B}{2}:{\text{cosec}}\frac{C}{2}$$
B.
$$\sin \frac{A}{2}:\sin \frac{B}{2}:\sin \frac{C}{2}$$
C.
$$\sec \frac{A}{2}:\sec \frac{B}{2}:\sec \frac{C}{2}$$
D.
None of these
Answer :
$${\text{cosec}}\frac{A}{2}:{\text{cosec}}\frac{B}{2}:{\text{cosec}}\frac{C}{2}$$
View Solution
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Here, $$BD : DC = c : b.$$
But $$BD + DC = a;$$
$$\therefore \,\,BD = \frac{c}{{b + c}} \cdot a.$$
$$\eqalign{
& {\text{In}}\,\,\vartriangle ABD,\frac{{BD}}{{\sin \frac{A}{2}}} = \frac{{AD}}{{\sin B}} \cr
& \therefore \,\,AD = \frac{{ca}}{{b + c}} \cdot \frac{{\sin B}}{{\sin \frac{A}{2}}} = \frac{{2\vartriangle }}{{b + c}}{\text{cosec}}\frac{A}{2}. \cr
& {\text{Also, }}\frac{{AI}}{{ID}} = \frac{{AB}}{{BD}} = \frac{c}{{\frac{{ca}}{{\left( {b + c} \right)}}}} = \frac{{b + c}}{a} \cr
& \Rightarrow \,\,AI = \frac{{b + c}}{{a + b + c}} \cdot AD = \frac{\vartriangle }{s}{\text{cosec}}\frac{A}{2}. \cr} $$
Similarly for $$BI$$ and $$CI.$$
42.
From the top of a cliff $$50\,m$$ high, the angles of depression of the top and bottom of a tower are observed to be $${30^ \circ }$$ and $${45^ \circ }.$$ The height of tower is
A.
$$50\,m$$
B.
$$50\sqrt 3 \,m$$
C.
$$50\left( {\sqrt 3 - 1} \right)m$$
D.
$$50\left( {1 - \frac{{\sqrt 3 }}{3}} \right)m$$
Answer :
$$50\left( {1 - \frac{{\sqrt 3 }}{3}} \right)m$$
View Solution
Discuss Question
Let height of the tower be $$h\,m$$ and distance between tower and cliffbe $$x\,m.$$
$$\eqalign{
& \therefore CD = h,BD = x \cr
& {\text{In }}\Delta \,ABD,\tan {45^ \circ } = \frac{{AB}}{{BD}} \cr
& {\text{or }}1 = \frac{{50}}{x} \cr
& x = 50\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
$$\eqalign{
& {\text{In }}\Delta \,AEC \cr
& \tan {30^ \circ } = \frac{{AE}}{{EC}} = \frac{{AB - EB}}{{EC}} = \frac{{AB - DC}}{{BD}}\left( {\because EB = DC,EC = BD} \right) \cr
& \frac{1}{{\sqrt 3 }} = \frac{{50 - h}}{x}{\text{ or }}x = 50\sqrt 3 - h\sqrt 3 \cr
& {\text{or }}50 = 50\sqrt 3 - h\sqrt 3 {\text{ or }}h\sqrt 3 = 50\sqrt 3 - 50 \cr
& {\text{or }}h = \frac{{50\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} = 50\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr
& \therefore h = 50\left( {1 - \frac{{\sqrt 3 }}{3}} \right) \cr} $$
43.
Let $${A_0},{A_1},{A_2},{A_3},{A_4}$$ and $${A_5}$$ be the consecutive vertices of a regular hexagon inscribed in a unit circle. The product of the lengths of $${A_0}{A_1},{A_0}{A_2}$$ and $${A_0}{A_4}$$ is
A.
$$\frac{3}{4}$$
B.
$${3\sqrt 3 }$$
C.
$$3$$
D.
$$\frac{{3\sqrt 3 }}{2}$$
Answer :
$$3$$
View Solution
Discuss Question
$$\eqalign{
& {A_0}{A_1} = 2 \times 1 \times \cos {60^ \circ } = 1 = {A_1}{A_2} \cr
& \cos {120^ \circ } = \frac{{{A_0}A_1^2 + {A_1}A_2^2 - {A_0}A_2^2}}{{2 \cdot {A_0}{A_1} \cdot {A_1}{A_2}}} = \frac{{1 + 1 - {A_0}A_2^2}}{{2 \cdot 1 \cdot 1}} \cr
& \therefore \,\,{A_0}{A_2} = \sqrt 3 = {A_0}{A_4} \cr
& \therefore \,\,{A_0}{A_1} \times {A_0}{A_2} \times {A_0}{A_4} = 1 \times \sqrt 3 \times \sqrt 3 = 3. \cr} $$
44.
In a $$\vartriangle ABC,\left( {c + a + b} \right)\left( {a + b - c} \right) = ab.$$ The measure of $$\angle C$$ is
A.
$$\frac{\pi }{3}$$
B.
$$\frac{\pi }{6}$$
C.
$$\frac{2\pi }{3}$$
D.
None of these
Answer :
$$\frac{2\pi }{3}$$
View Solution
Discuss Question
$$\eqalign{
& 2s\left( {2s - 2c} \right) = ab\,\,\,{\text{or, }}\frac{{s\left( {s - c} \right)}}{{ab}} = \frac{1}{4} \cr
& {\text{or, }}{\cos ^2}\frac{C}{2} = \frac{1}{4}\,\,\,{\text{or,}}\,\,\cos \frac{C}{2} = \frac{1}{2}\,\,\,\left( {\because \,\,\,\frac{C}{2}\,{\text{must be acute}}} \right). \cr} $$
45.
In a $$\vartriangle ABC,\cos A + \cos B + \cos C > 1$$ only if the triangle is
A.
acute angled
B.
obtuse angled
C.
right angled
D.
the nature of the triangle cannot be determined
Answer :
the nature of the triangle cannot be determined
View Solution
Discuss Question
$$\eqalign{
& \cos A + \cos B + \cos C > 1 \cr
& \Rightarrow \,\,1 + 4\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 1 \cr
& \therefore \,\,\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 0 \cr
& \Rightarrow \,\,\frac{A}{2},\frac{B}{2},\frac{C}{2}\,\,{\text{are acute}}{\text{.}} \cr} $$
46.
The number of possible triangles $$ABC$$ in which $$BC = \sqrt {11} \,cm,CA = \sqrt {13} \,cm$$ and $$A = {60^ \circ }$$ is
A.
0
B.
1
C.
2
D.
None of these
Answer :
2
View Solution
Discuss Question
Using, $$\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}},\frac{1}{2} = \frac{{13 + {c^2} - 11}}{{2 \cdot \sqrt {13} \cdot c}}$$
So, $${c^2} = \sqrt {13} \cdot c + 2 = 0.$$ This gives two values of $$c.$$
47.
The horizontal distance between two towers is 60 metres and the angular depression of the top of the first tower as seen from the top of the second. is $${30^ \circ }.$$ If the height of the second tower be 150 metres, then the height of the first tower is
A.
$$150 - 60\sqrt 3 \,m$$
B.
$$90\,m$$
C.
$$150 - 20\sqrt 3 \,m$$
D.
None of these
Answer :
$$150 - 20\sqrt 3 \,m$$
View Solution
Discuss Question
$$\eqalign{
& \left( {150 - h} \right)\cot {30^ \circ } = 60 \cr
& \Rightarrow h = 150 - 20\sqrt 3 \,m \cr} $$
48.
In a $$\vartriangle ABC,2s = $$ perimeter and $$R =$$ circumradius. Then $$\frac{s}{R}$$ is equal to
A.
$$\sin A + \sin B + \sin C$$
B.
$$\cos A + \cos B + \cos C$$
C.
$$\sin \frac{A}{2} + \sin \frac{B}{2} + \sin \frac{C}{2}$$
D.
None of these
Answer :
$$\sin A + \sin B + \sin C$$
View Solution
Discuss Question
$$\frac{s}{R} = \frac{{\left( {a + b + c} \right)}}{{2R}} = \frac{a}{{2R}} + \frac{b}{{2R}} + \frac{c}{{2R}} = \sin A + \sin B + \sin C.$$
49.
In a triangle the sum of two sides is $$x$$ and the product of the same sides is $$y.$$ If $${x^2} - {c^2} = y,$$ where $$c$$ is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is
A.
$$\frac{{3y}}{{2x\left( {x + c} \right)}}$$
B.
$$\frac{{3y}}{{2c\left( {x + c} \right)}}$$
C.
$$\frac{{3y}}{{4x\left( {x + c} \right)}}$$
D.
$$\frac{{3y}}{{4c\left( {x + c} \right)}}$$
Answer :
$$\frac{{3y}}{{2c\left( {x + c} \right)}}$$
View Solution
Discuss Question
Let two sides of $$\Delta $$ be $$a$$ and $$b.$$
Then $$a + b = x$$ and $$ab = y$$
Also given $${x^2} - {c^2} = y,$$ where $$c$$ is the third side of $$\Delta .$$
$$\eqalign{
& \Rightarrow \,\,{\left( {a + b} \right)^2} - {c^2} = ab \cr
& \Rightarrow \,\,{a^2} + {b^2} - {c^2} = - ab \cr
& \Rightarrow \,\,\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} = - \frac{1}{2} \cr
& \Rightarrow \,\,\cos c = - \frac{1}{2} \cr
& \Rightarrow \,\,c = {120^ \circ } \cr} $$
$$\therefore \,\,\frac{r}{R} = \frac{\Delta }{s} \times \frac{{4\Delta }}{{abc}}$$ where $$\Delta =\,\,$$ area of triangle
$$\eqalign{
& \Rightarrow \,\,\frac{r}{R} = \frac{{4{\Delta ^2}}}{{\frac{{\left( {a + b + c} \right)}}{2}abc}} \cr
& = \frac{{8 \times {{\left( {\frac{1}{2}ab\sin c} \right)}^2}}}{{\left( {a + b + c} \right)abc}} \cr
& \Rightarrow \,\,\frac{{2{a^2}{b^2}{{\sin }^2}{{120}^ \circ }}}{{\left( {a + b + c} \right)abc}} \cr
& = \frac{{2ab \times \frac{3}{4}}}{{\left( {x + c} \right)c}} \cr
& = \frac{{3y}}{{2c\left( {x + c} \right)}} \cr} $$
50.
If in a $$\vartriangle ABC,2\cos A\sin C = \sin B$$ then the triangle is
A.
equilateral
B.
isosceles
C.
right angled
D.
None of these
Answer :
isosceles
View Solution
Discuss Question
$$\eqalign{
& 2\cos A \cdot \sin \,C = \sin \left( {C + A} \right) = \sin \,C \cdot \cos A + \cos \,C \cdot \sin \,A \cr
& \Rightarrow \,\,\sin \left( {C - A} \right) = 0 \cr
& \Rightarrow \,\,C = A. \cr} $$