
In \[\Delta ABC\], if \[\angle C = {90^ \circ }\], \[\angle A = {30^ \circ }\], c = 20, then find the value of a and b.
A. 10,10
B. 10, \[10\sqrt 3 \]
C. 5, \[5\sqrt 3 \]
D. 8, \[8\sqrt 3 \]
Answer
163.2k+ views
Hint: First we will find \[\angle B\] by using the sum of interior angles of a triangle. To solve the question, we will use sine law. By using sine law, we will find the value of a and b.
Formula used:
Sine Law:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Complete step by step solution:
We know that the sum of interior angles of a triangle is \[{180^ \circ }\].
Thus, \[\angle A + \angle B + \angle C = {180^ \circ }\]
Substitute \[\angle C = {90^ \circ }\], \[\angle A = {30^ \circ }\]
\[{90^ \circ } + \angle B + {30^ \circ } = {180^ \circ }\]
\[ \Rightarrow {120^ \circ } + \angle B = {180^ \circ }\]
\[ \Rightarrow \angle B = {180^ \circ } - {120^ \circ }\]
\[ \Rightarrow \angle B = {60^ \circ }\]
We know the sine law,
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Now substitute \[\angle C = {90^ \circ }\], \[\angle A = {30^ \circ }\],\[\angle B = {60^ \circ }\], c = 20 in the sine law:
\[\dfrac{a}{{\sin {{30}^ \circ }}} = \dfrac{b}{{\sin {{60}^ \circ }}} = \dfrac{{20}}{{\sin {{90}^ \circ }}}\]
Substitute the value of \[\sin {30^ \circ }\], \[\sin {60^ \circ }\], \[\sin {90^ \circ }\]
\[ \Rightarrow \dfrac{a}{{\dfrac{1}{2}}} = \dfrac{b}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{20}}{1}\]
Therefore,
\[ \Rightarrow \dfrac{a}{{\dfrac{1}{2}}} = \dfrac{{20}}{1}\] and \[ \Rightarrow \dfrac{b}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{20}}{1}\]
\[ \Rightarrow a = \dfrac{1}{2} \times \dfrac{{20}}{1}\] and \[ \Rightarrow b = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{20}}{1}\]
\[ \Rightarrow a = 10\] and \[ \Rightarrow b = 10\sqrt 3 \]
Hence option B is the correct option.
Note: It can be solved by using another method. The angles of the triangle are \[{30^ \circ },{60^ \circ },{90^ \circ }\]. The ratio of the sine of the angle is \[\dfrac{1}{2}:\dfrac{{\sqrt 3 }}{2}:1 = 1:\sqrt 3 :2\]. The sides of the triangle will be x, \[\sqrt 3 x\], \[2x\]. Since c = 20 \[ \Rightarrow 2x = 20 \Rightarrow x = 10\]. The rest sides of the triangle are 10 and \[10\sqrt 3 \].
Formula used:
Sine Law:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Complete step by step solution:
We know that the sum of interior angles of a triangle is \[{180^ \circ }\].
Thus, \[\angle A + \angle B + \angle C = {180^ \circ }\]
Substitute \[\angle C = {90^ \circ }\], \[\angle A = {30^ \circ }\]
\[{90^ \circ } + \angle B + {30^ \circ } = {180^ \circ }\]
\[ \Rightarrow {120^ \circ } + \angle B = {180^ \circ }\]
\[ \Rightarrow \angle B = {180^ \circ } - {120^ \circ }\]
\[ \Rightarrow \angle B = {60^ \circ }\]
We know the sine law,
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Now substitute \[\angle C = {90^ \circ }\], \[\angle A = {30^ \circ }\],\[\angle B = {60^ \circ }\], c = 20 in the sine law:
\[\dfrac{a}{{\sin {{30}^ \circ }}} = \dfrac{b}{{\sin {{60}^ \circ }}} = \dfrac{{20}}{{\sin {{90}^ \circ }}}\]
Substitute the value of \[\sin {30^ \circ }\], \[\sin {60^ \circ }\], \[\sin {90^ \circ }\]
\[ \Rightarrow \dfrac{a}{{\dfrac{1}{2}}} = \dfrac{b}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{20}}{1}\]
Therefore,
\[ \Rightarrow \dfrac{a}{{\dfrac{1}{2}}} = \dfrac{{20}}{1}\] and \[ \Rightarrow \dfrac{b}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{{20}}{1}\]
\[ \Rightarrow a = \dfrac{1}{2} \times \dfrac{{20}}{1}\] and \[ \Rightarrow b = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{20}}{1}\]
\[ \Rightarrow a = 10\] and \[ \Rightarrow b = 10\sqrt 3 \]
Hence option B is the correct option.
Note: It can be solved by using another method. The angles of the triangle are \[{30^ \circ },{60^ \circ },{90^ \circ }\]. The ratio of the sine of the angle is \[\dfrac{1}{2}:\dfrac{{\sqrt 3 }}{2}:1 = 1:\sqrt 3 :2\]. The sides of the triangle will be x, \[\sqrt 3 x\], \[2x\]. Since c = 20 \[ \Rightarrow 2x = 20 \Rightarrow x = 10\]. The rest sides of the triangle are 10 and \[10\sqrt 3 \].
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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
