
If in a \[\Delta ABC\],\[\angle A = 45^\circ \],\[\angle C = 60^\circ \], then \[a + c\sqrt 2 \]
A. \[b\]
B. \[2b\]
C. \[\sqrt {2b} \]
D. \[\sqrt {3b} \]
Answer
163.5k+ views
Hint
In this case we are given the values of some angles of like \[\angle A = {45^\circ },\angle C = {60^\circ }\] and asked to determine the value of \[a + c\sqrt 2 \] and we will use the extended sine rule to determine the relationship between the length of the triangle's sides and its circumradius to obtain the desired result.
Formula used:
Sine rule formula:
\[\frac{a}{{sinA}} = \frac{b}{{sinB}} = \frac{c}{{sinC}} = 2R\]
Complete step-by-step solution:
The given angle is \[A = 45^\circ \], \[C = 60^\circ \]
\[A + B + C = \pi \]
By substituting the values on the equation, it becomes
\[ = > B = 75^\circ \]
\[a + c\sqrt 2 = k\sin A + k\sin C(\sqrt 2 )\]
\[ = 2k(\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }})\]
The values on the equation becomes,
\[ = 2k\sin 75^\circ \]
Then, the equation becomes
\[ = 2k\sin B\]
\[a + c\sqrt 2 = 2b\]
So, option B is correct.
Note
You need to first determine the lengths of ABC in order to solve this problem. The lengths of AB and BC are \[2\] and \[3\], respectively. As a result, dragging a downward will result in the intersection of AD and BC.
Two lines are said to intersect when they have exactly one point in common. There is a point at which the intersecting lines meet. The point of intersection is the same location that appears on all intersecting lines. There will be a place where the two coplanar, non-parallel straight lines intersect. Here, point O, the intersection point, is where lines A and B meet.
In this case we are given the values of some angles of like \[\angle A = {45^\circ },\angle C = {60^\circ }\] and asked to determine the value of \[a + c\sqrt 2 \] and we will use the extended sine rule to determine the relationship between the length of the triangle's sides and its circumradius to obtain the desired result.
Formula used:
Sine rule formula:
\[\frac{a}{{sinA}} = \frac{b}{{sinB}} = \frac{c}{{sinC}} = 2R\]
Complete step-by-step solution:
The given angle is \[A = 45^\circ \], \[C = 60^\circ \]
\[A + B + C = \pi \]
By substituting the values on the equation, it becomes
\[ = > B = 75^\circ \]
\[a + c\sqrt 2 = k\sin A + k\sin C(\sqrt 2 )\]
\[ = 2k(\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }})\]
The values on the equation becomes,
\[ = 2k\sin 75^\circ \]
Then, the equation becomes
\[ = 2k\sin B\]
\[a + c\sqrt 2 = 2b\]
So, option B is correct.
Note
You need to first determine the lengths of ABC in order to solve this problem. The lengths of AB and BC are \[2\] and \[3\], respectively. As a result, dragging a downward will result in the intersection of AD and BC.
Two lines are said to intersect when they have exactly one point in common. There is a point at which the intersecting lines meet. The point of intersection is the same location that appears on all intersecting lines. There will be a place where the two coplanar, non-parallel straight lines intersect. Here, point O, the intersection point, is where lines A and B meet.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
