
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 \[10\sqrt 3 \].
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