Free Objective Test 01 Practice Test - 11th and 12th

Question 1

Number of triangles possible for a given b, c and B(acute angle) under the condition that b < c sin B. Where b, c are the sides and B is the angle opposite to b.

A. 0

B. 1

C. 2

D. 3

SOLUTION

Solution : A

Let’s construct the lines and angles which constitute the triangle.

We can see once c and B are fixed the least value it should have in order to form a triangle is b = c sin B. In fact, you can form 2 triangles if b > c sin B and B is acute. If B is obtuse by default b > c $\Rightarrow$ b > c sin B. Therefore no triangle is possible when b < c sin B irrespective of B being acute or obtuse. Answer is 0 triangles are possible.

Question 2

Given angle $A = 60^{\circ}, c = \sqrt{3} - 1, b = \sqrt{3} + 1$ . Solve the triangle

$a = \sqrt{6}, B = 15^{\circ}, C = 100^{\circ}$

$a = \sqrt{6}, B = 15^{\circ}, C = 105^{\circ}$

$a = \sqrt{12}, B = 15^{\circ}, C = 105^{\circ}$

$a = \sqrt{6}, B = 105^{\circ}, C = 15^{\circ}$

SOLUTION

Solution : D

$A = 60^{\circ} \Rightarrow B + C = 120^{\circ} W e k n o w, t a n \frac{B - C}{2} = \frac{b - c}{b + c} c o t \frac{A}{2} \Rightarrow t a n \frac{B - C}{2} = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{(\sqrt{3} + 1) + (\sqrt{3} - 1)} c o t (\frac{60}{2}) = \frac{2}{2 \sqrt{3}} (\sqrt{3}) = 1 ∴ \frac{B - C}{2} = 45^{\circ} \Rightarrow B - C = 90^{\circ} B + C = 120^{\circ} ∴ B = 105^{\circ} a n d C = 15^{\circ}, F r o m s i n e r u l e, a = \frac{b s i n A}{s i n B} = \sqrt{3} + 1 (\frac{s i n 60}{s i n 105}) s i n 105 = s i n (60^{\circ} + 45^{\circ}) = \frac{\sqrt{3}}{2} (\frac{1}{\sqrt{2}}) + \frac{1}{2} (\frac{1}{\sqrt{2}}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}} a = \frac{(\sqrt{3} + 1) (\sqrt{3}) (2) (\sqrt{2})}{(\sqrt{3} + 1) 2} = \sqrt{6} ∴ a = \sqrt{6}, B = 105^{\circ}, C = 15^{\circ}$
Option d is correct.

Question 3

In a triangle ABC, a = 3, b = 5, c = 7. Find the angle opposite to C.

$60^{\circ}$

$90^{\circ}$

$120^{\circ}$

$150^{\circ}$

SOLUTION

Solution : C

We know from trigonometric ratios of half angles that $t a n \frac{A}{2} = \sqrt{\frac{(s - b) (s - c)}{(s - a) (s)}}$
Here we need angle opposite to c i.e., C.
Semiperimeter, $S = \frac{a + b + c}{2} = \frac{3 + 5 + 7}{2} = \frac{15}{2}$
$t a n \frac{C}{2} = \sqrt{\frac{(s - b) (s - a)}{(s - c) (s)}}$
$= \sqrt{\frac{(s - 3) (s - 5)}{(s - 7) s}} = \sqrt{\frac{(15 - 6) (15 - 10) (2)}{4 (15 - 14) (\frac{15}{2})}}$
$= \sqrt{\frac{9 (5)}{15}} = \sqrt{3} \Rightarrow \frac{C}{2} = 60^{\circ} \Rightarrow C = 120^{\circ}$

Question 4

In a triangle $a^{2} + b^{2} + c^{2} = c a + a b \sqrt{3}$ , then triangle is

A. Equilateral

B. Right angled and isosceles

C. Right angled

D. None of these

SOLUTION

Solution : C

Given, $a^{2} + b^{2} + c^{2} = c a + a b \sqrt{3} a^{2} + b^{2} + c^{2} - c a - a b \sqrt{3} = 0 \Rightarrow {(\frac{a \sqrt{3}}{2} - b)}^{2} + {(\frac{a}{2} - c)}^{2} = 0 \Rightarrow a = \frac{2 b}{\sqrt{3}} a n d a = \frac{c}{2} W e c a n s e e, a^{2} + b^{2} = c^{2} s i n B = \frac{b}{a} = \frac{\sqrt{3}}{2} \neq 1 \Rightarrow N o t i s o s c e l e s ∴ G i v e n t r i a n g l e i s a R i g h t a n g l e d t r i a n g l e w h i c h i s n o t i s o s c e l e s$ .
Option C is correct.

Question 5

In a triangle ABC if a =13, b = 8 and c = 7, then find sin A.

\frac{1}{2}

\frac{\sqrt{3}}{2}

\frac{\sqrt{3}}{4}

\frac{1}{4}

SOLUTION

Solution : B

In trigonometry one problem can be solved in multiple ways but it is wise to use the best way to get the answer or at least not to use the lengthiest way. Here, we can apply sine rule and get the answer but have a look at the problem again. We are given all the sides. And when sides are given and angles are asked to find ,using cosine rule is a better option.
According to cosine rule -
$c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$c o s A = \frac{(8)^{2} + (7)^{2} - (13)^{2}}{2 (8) (7)}$ (On putting values of a,b & c)
$c o s A = - \frac{1}{2}$
$A = \frac{2 π}{3}$
$s i n A = s i n \frac{2 π}{3} = \frac{\sqrt{3}}{2}$

Question 6

Which of the following relations hold true ?

r = \frac{Δ}{s}, r = 2 R s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

r = \frac{Δ}{s}, r = R s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

r = \frac{Δ}{s}, r = 2 R s i n \frac{A}{3} s i n \frac{B}{3} s i n \frac{C}{3}

r = \frac{Δ}{s}, r = 2 R s i n \frac{A}{3} s i n \frac{B}{2} s i n \frac{C}{2}

SOLUTION

Solution : A

Relation between in radius, Area of triangle and semiperimeter can be given by $r = \frac{Δ}{s}$ and relation between inradius and circumradius can be given by $r = 2 R s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}$

Question 7

Which of the following holds true in a Right angled triangle?

A. r + R = s

B. r + 2R = s

C. 2r + R = s

D. r + R = 2s

SOLUTION

Solution : B

We know, for a Right angled triangle,
$R = \frac{a}{2} w h e r e A = 90^{\circ} A l s o, r = (s - a) t a n \frac{A}{2} = (s - a) t a n 45^{\circ} = (s - a) = s - 2 R \Rightarrow r + 2 R = s$

Question 8

Relation between Exradius ,semiperimeter and circumradius can be given by
( $r_{1}$ is the radius of the circle opposite the angle A)

r_{1} = \frac{2 Δ}{s - a}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

r_{1} = \frac{Δ}{s - a}, r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}

r_{1} = \frac{2 Δ}{(s - a) (s - b)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

r_{1} = \frac{Δ (s - a)}{(s - b) (s - c)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

SOLUTION

Solution : B

Relation between Exradius and semiperimeter can be given by, $r_{1} = \frac{Δ}{s - a}$ where $r_{1}$ is the radius of the circle opposite the angle A. and it's helpful to remember the identity $r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}$

Question 9

In a $Δ A B C, a (b c o s C - c c o s B)$ is equal to -

A. b + c

B. b - c

b^{2} - c^{2}

b^{2} + c^{2}

SOLUTION

Solution : C

We know from cosine rule
$c o s C = \frac{a^{2} + b^{2} - C^{2}}{2 a b}$
$c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$
So we'll use these in the given expression.
$a {b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) - c (\frac{a^{2} + c^{2} - b^{2}}{2 a c})}$
$\frac{a^{2} + b^{2} - c^{2} - a^{2} - c^{2} + b^{2}}{2}$
$= \frac{2 (b^{2} - c^{2})}{2}$
$= b^{2} - c^{2}$

Question 10

If the median AD of a triangle ABC is perpendicular to BC, then which of the following is true always?

A. B = C

B. A = C

C. A = B

D. A = B = C

SOLUTION

Solution : A

As the median is perpendicular by SAS similarity angle B should be equal to angle C.

Question 11

If in a $Δ$ ABC, perimeter is 12 units, b = 3 units, c = 5 units then find $s i n (\frac{A}{2})$ .

\frac{1}{\sqrt{5}}

\sqrt{5}

\frac{1}{\sqrt{3}}

\sqrt{3}

SOLUTION

Solution : A

We know, $s i n (\frac{A}{2}) = \sqrt{\frac{(s - b) (s - c)}{b . c}}$
Where"s" is the semiperimeter and b & c are the sides of a triangle.
We'll do the appropriate substitution -
$s i n (\frac{A}{2}) = \sqrt{\frac{(6 - 3) (6 - 5)}{3.5}}$
$s i n (\frac{A}{2}) = \frac{1}{\sqrt{5}}$

Question 12

Find the area of a triangle if two sides are of 5 units, 8 units and the angle between them is $30^{\circ}$ .

A. 10

B. 20

C. 30

D. 50

SOLUTION

Solution : A

We know that the area of a triangle = $\frac{1}{2}$ a.b.sin C
In this C is the angle opposite to the side of length "c" or made in between sides a & b.
Area = $\frac{1}{2} 5.8. s i n 30^{\circ}$
Area = 10 units.

Question 13

If sides of a triangle are 3,5,7 then the area of that triangle is -

\sqrt{42.18}

\sqrt{62.78}

\sqrt{97.89}

\sqrt{23.12}

SOLUTION

Solution : A

We know the area of a triangle can be calculated using its sides through Heron's formula
$\sqrt{s (s - a) (s - b) (s - c)}$
Where "s" is the semiperimeter of the triangle.
And a, b, c are the sides of it.
So, $s = \frac{3 + 5 + 7}{2} = 7.5$
And Area = $\sqrt{7.5 (7.5 - 3) (7.5 - 5) (7.5 - 7)}$
$= \sqrt{42.18}$

Question 14

In an equilateral triangle sum of the distances of orthocenter from all the vertices is equal to -
(Where 'R' is the circumradius of triangle)

A. R

D. 4R

SOLUTION

Solution : C

Let the triangle be ABC and the orthocenter be O.
We know that OA = 2RcosA

It is given that triangle is equilateral. So cosA = cos (60°) = $\frac{1}{2}$
And OA = R
Similarly, OB = OC = R
And the sum of them = OA + OB + OC = 3R

Question 15

In the following figure, the area of the pedal triangle is -

\frac{1}{2} R^{2} . sin A.sin B.sin C

\frac{1}{2} R^{2} . sin 2A.sin 2B.sin 2C

\frac{1}{2} R^{2} . cos A.cos B.cos C

D. None of these

SOLUTION

Solution : B

Since, area of a triangle
$= \frac{1}{2} (product of the sides) \times sine of the included angle)$
$= \frac{1}{2} R sin 2 B \cdot sin (180 ° - 2 A)$
$= \frac{1}{2} R^{2} sin 2 A \cdot sin 2 B \cdot sin 2 C$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsObjective Test 02Objective Test 01Objective Test 02

Related Exams
Objective Test 02
Objective Test 01
Objective Test 02