1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

With the usual notation, in $$\Delta $$ABC, if $$\angle A + \angle B$$ = 120o, a = $$\sqrt 3 $$ $$+$$ 1, b = $$\sqrt 3 $$ $$-$$ 1 then the ratio $$\angle A:\angle B,$$ is -
A
9 : 7
B
7 : 1
C
5 : 3
D
3 : 1

Explanation

A + B = 120o


$$\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)$$

$$ = {{\sqrt 3 + 1 - \sqrt 3 + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3 = 1$$

$${{A - B} \over 2} = {45^o}$$

$$ \Rightarrow A - B = {90^o}$$

$$ \ A + B = {120^o}$$

$$2A = {210^o}$$

$$A = {105^o}$$

$$B = {15^o}$$

$$ \therefore $$ $$\angle A:\angle B,$$ = 7 : 1
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 – c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :
A
$${y \over {\sqrt 3 }}$$
B
$${c \over 3}$$
C
$${c \over {\sqrt 3 }}$$
D
$${3 \over 2}$$y

Explanation

Given a + b = x and ab = y

If x2 $$-$$ c2 = y $$ \Rightarrow $$ (a + b)2 $$-$$ c2 = ab

$$ \Rightarrow $$  a2 + b2 $$-$$ c2 = $$-$$ ab

$$ \Rightarrow $$   $${{{a^2} + {b^2} - {c^2}} \over {2ab}} = - {1 \over 2}$$

$$ \Rightarrow \cos C = - {1 \over 2}$$

$$ \Rightarrow \angle C = {{2\pi } \over 3}$$

$$R = {c \over {2\sin C}} = {c \over {\sqrt 3 }}$$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Evening Slot

Given $${{b + c} \over {11}} = {{c + a} \over {12}} = {{a + b} \over {13}}$$ for a $$\Delta $$ABC with usual notation.

If   $${{\cos A} \over \alpha } = {{\cos B} \over \beta } = {{\cos C} \over \gamma },$$ then the ordered triad ($$\alpha $$, $$\beta $$, $$\gamma $$) has a value
A
(19, 7, 25)
B
(7, 19, 25)
C
(5, 12, 13)
D
(3, 4, 5)

Explanation

b + c = 11$$\lambda $$, c + a = 12$$\lambda $$, a + b = 13$$\lambda $$

$$ \Rightarrow $$  a = 7$$\lambda $$, b = 6$$\lambda $$, c = 5$$\lambda $$

(using cosine formula)

cosA = $${1 \over 5},$$ cosB = $${19 \over 35},$$ cosC = $${5 \over 7},$$

$$\alpha $$ : $$\beta $$ : $$\gamma $$ $$ \Rightarrow $$  7 : 19 : 25
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 12th January Evening Slot

If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30o and the angle of depression of reflection of the cloud in the lake from P be 60o, then the height of the cloud (in meters) from the surface of the lake is :
A
45
B
42
C
50
D
60

Explanation


tan 30o = $${x \over y} \Rightarrow y = \sqrt 3 x\,\,\,\,....(i)$$

tan 60o = $${{25 + x + 25} \over y}$$

$$ \Rightarrow \,\,\sqrt 3 y = 50 + x$$

$$ \Rightarrow $$  $$3x = 50 + x$$

$$ \Rightarrow $$   x = 25 m

$$ \therefore $$   Height of cloud from surface

= 25 + 25 = 50m

Questions Asked from Properties of Triangle

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