Solution Of Triangles(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

SOLUTION OF TRIANGLES MCQs

Total Questions : 15 | Page 1 of 2 pages

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.

Discuss Question

Answer: Option A. -> 0
:
A
Let’s construct the lines and angles which constitute the triangle.

Number Of Triangles Possible For A Given B, C And B(acute An...

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

$a = \sqrt{6}, B = 15^{\circ}, C = 100^{\circ}$
$a = \sqrt{6}, B = 15^{\circ}, C = 105^{\circ}$
$a = \sqrt{12}, B = 15^{\circ}, C = 105^{\circ}$
$a = \sqrt{6}, B = 105^{\circ}, C = 15^{\circ}$

Discuss Question

Answer: Option D. ->

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

:
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.

$60^{\circ}$
$90^{\circ}$
$120^{\circ}$
$150^{\circ}$

Discuss Question

Answer: Option C. ->

120^{\circ}

:
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

Equilateral
Right angled and isosceles
Right angled
None of these

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Answer: Option C. -> Right angled
:
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.

Which of the following relations hold true ?

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

Discuss Question

Answer: Option A. ->

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}

:
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 6.

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

$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{4}$
$\frac{1}{4}$

Discuss Question

Answer: Option B. ->

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

:
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 7.

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

r + R = s
r + 2R = s
2r + R = s
r + R = 2s

Discuss Question

Answer: Option B. -> r + 2R = s
:
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)

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