
Sides of a triangle are \[2\;{\rm{cm}},\sqrt 6 \;{\rm{cm,}}\]and\[\left( {\sqrt 3 + 1} \right){\rm{cm}}\]. The smallest angle of the triangle is
A. \[30^\circ \]
B. \[45^\circ \]
C. \[60^\circ \]
D. \[75^\circ \]
Answer
163.2k+ views
Hint:
The triangle's lengths, widths, and heights can be used to calculate the sides of a triangle, although this solution isn't always accurate. In rare circumstances, one side may be wider or shorter than the other.
Use of fundamental geometric concepts like proportionality and likeness is crucial when resolving issues like these. You can use these ideas to work out which measures connect to which triangle sides and how they relate to one another. Once these relationships have been established, include them into your equation to allow you to solve for the required sides.
Complete step-by-step solution:
Given that \[a = 2,b = \sqrt 6 ,c = \sqrt 3 + 1\]
\[\cos A = \frac{{6 + 3 + 1 + 2\sqrt 3 - 4}}{{2\sqrt 6 (\sqrt 3 + 1)}}\]
Calculate the sum or difference
\[\cos A = \frac{{6 + 2\sqrt 3 }}{{2\sqrt 6 (\sqrt 3 + 1)}}\]
Factor out 2 from the expression
\[\cos A = \frac{{2\left( {3 + \sqrt 3 } \right)}}{{2\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Cancel out the common factor 2
\[\cos A = \frac{{\left( {3 + \sqrt 3 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Rationalize the denominator
\[\cos A = \frac{{\left( {3 + \sqrt 3 } \right)\sqrt 6 }}{{6\left( {\sqrt 3 + 1} \right)}}\]
Distribute \[\sqrt 6 \]through the parentheses
\[\cos A = \frac{{3 + \sqrt 6 \sqrt {18} }}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Simplify the radical expression
\[\cos A = \frac{{3 + \sqrt 6 + 3 + \sqrt 2 }}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Factor out 3 from the expression
\[\cos A = \frac{{3\left( {\sqrt 6 + \sqrt 2 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Factor out \[\sqrt 2 \] from the expression
\[\cos A = \frac{{\sqrt 2 \left( {\sqrt 3 + 1} \right)}}{{2\left( {\sqrt 3 + 1} \right)}}\]
Cancel out the common factor \[\sqrt 3 + 1\]
\[\cos A = \frac{{\sqrt 2 }}{2}\]
Using the unit circle, find the angles for which the cosine equals \[\frac{{\sqrt 2 }}{2}\]
As a result, \[A = {45^\circ }\].
So, Option B is correct.
Note:
Students must consider the length and width of each side in order to solve this issue. Since \[\left( {\sqrt 3 + 1} \right)\]cm is shorter than \[2cm\], the shorter of the two sides must be the other. Furthermore, it must be the smaller side because it is \[\sqrt 6 cm\] wider than both sides.
As students try to determine the angle that is smaller than \[45^\circ \], they may mistakenly assume \[72^\circ \] since the triangle is an equal right triangle with two separate sides.
The triangle's lengths, widths, and heights can be used to calculate the sides of a triangle, although this solution isn't always accurate. In rare circumstances, one side may be wider or shorter than the other.
Use of fundamental geometric concepts like proportionality and likeness is crucial when resolving issues like these. You can use these ideas to work out which measures connect to which triangle sides and how they relate to one another. Once these relationships have been established, include them into your equation to allow you to solve for the required sides.
Complete step-by-step solution:
Given that \[a = 2,b = \sqrt 6 ,c = \sqrt 3 + 1\]
\[\cos A = \frac{{6 + 3 + 1 + 2\sqrt 3 - 4}}{{2\sqrt 6 (\sqrt 3 + 1)}}\]
Calculate the sum or difference
\[\cos A = \frac{{6 + 2\sqrt 3 }}{{2\sqrt 6 (\sqrt 3 + 1)}}\]
Factor out 2 from the expression
\[\cos A = \frac{{2\left( {3 + \sqrt 3 } \right)}}{{2\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Cancel out the common factor 2
\[\cos A = \frac{{\left( {3 + \sqrt 3 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Rationalize the denominator
\[\cos A = \frac{{\left( {3 + \sqrt 3 } \right)\sqrt 6 }}{{6\left( {\sqrt 3 + 1} \right)}}\]
Distribute \[\sqrt 6 \]through the parentheses
\[\cos A = \frac{{3 + \sqrt 6 \sqrt {18} }}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Simplify the radical expression
\[\cos A = \frac{{3 + \sqrt 6 + 3 + \sqrt 2 }}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Factor out 3 from the expression
\[\cos A = \frac{{3\left( {\sqrt 6 + \sqrt 2 } \right)}}{{\sqrt 6 \left( {\sqrt 3 + 1} \right)}}\]
Factor out \[\sqrt 2 \] from the expression
\[\cos A = \frac{{\sqrt 2 \left( {\sqrt 3 + 1} \right)}}{{2\left( {\sqrt 3 + 1} \right)}}\]
Cancel out the common factor \[\sqrt 3 + 1\]
\[\cos A = \frac{{\sqrt 2 }}{2}\]
Using the unit circle, find the angles for which the cosine equals \[\frac{{\sqrt 2 }}{2}\]
As a result, \[A = {45^\circ }\].
So, Option B is correct.
Note:
Students must consider the length and width of each side in order to solve this issue. Since \[\left( {\sqrt 3 + 1} \right)\]cm is shorter than \[2cm\], the shorter of the two sides must be the other. Furthermore, it must be the smaller side because it is \[\sqrt 6 cm\] wider than both sides.
As students try to determine the angle that is smaller than \[45^\circ \], they may mistakenly assume \[72^\circ \] since the triangle is an equal right triangle with two separate sides.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
