
Find the area of the triangle \[ABC\], in which \[a = 1,b = 2\] and \[\angle C = {60^o}\].
A. \[\dfrac{1}{2}\]
B. \[\sqrt 3 \]
C. \[\dfrac{{\sqrt 3 }}{2}\]
D. \[\dfrac{3}{2}\]
Answer
162k+ views
Hint:
In the given question, we need to find the area of \[\Delta ABC\]. For this, we need to use the following formula. After simplification of this, we will get the desired result.
Formula Used:
Suppose \[a, b\] and \[c\] are the sides and also \[A, B\] and \[C\] are the angles of a triangle \[ABC\] then,
Area of triangle\[ABC\] = \[\dfrac{1}{2}ab\sin C\]
\[\sin {60^o} = \dfrac{{\sqrt 3 }}{2}\]
Complete step-by-step answer:
Let \[a, b\] and \[c\] are the sides and \[A, B\] and \[C\] are the angles of a triangle \[ABC\].
Given that: \[a = 1, b = 2\] and \[\angle C = {60^o}\]
Thus,
Area of \[\Delta ABC = \dfrac{1}{2}ab\sin C\]
\[\Rightarrow\] Area of \[ \Delta ABC = \dfrac{1}{2} \times 1 \times 2\left( {\sin {{60}^o}} \right)\]
By simplifying, we get
\[ \Rightarrow\] Area of \[ \Delta ABC = 1 \times \left( {\sin {{60}^o}} \right)\]
\[ \Rightarrow\] Area of \[\Delta ABC = 1 \times \left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow\] Area of \[\Delta ABC = \left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
Hence, the area of a triangle \[ABC\] is \[\dfrac{{\sqrt 3 }}{2}\].
Therefore, the correct option is (C).
Additional Information : The total space occupied by the three sides of every specific triangle is known as its area. The fundamental method of calculating the area of a triangle is half the product of its base and height. It is possible to find the area of a triangle in different ways. We can also find the area of a triangle in two different ways such as \[ A(\Delta ABC) = \dfrac{1}{2}bc\sin A\] and \[ A(\Delta ABC) = \dfrac{1}{2}ac\sin B\] .
Note:
Many students make mistakes in writing the formula of area of a triangle in terms of sine. This is the only way through which we can solve the example in the simplest way. Also, it is essential to simplify carefully and write the proper value of trigonometric angle to get the desired result.
In the given question, we need to find the area of \[\Delta ABC\]. For this, we need to use the following formula. After simplification of this, we will get the desired result.
Formula Used:
Suppose \[a, b\] and \[c\] are the sides and also \[A, B\] and \[C\] are the angles of a triangle \[ABC\] then,
Area of triangle\[ABC\] = \[\dfrac{1}{2}ab\sin C\]
\[\sin {60^o} = \dfrac{{\sqrt 3 }}{2}\]
Complete step-by-step answer:
Let \[a, b\] and \[c\] are the sides and \[A, B\] and \[C\] are the angles of a triangle \[ABC\].
Given that: \[a = 1, b = 2\] and \[\angle C = {60^o}\]
Thus,
Area of \[\Delta ABC = \dfrac{1}{2}ab\sin C\]
\[\Rightarrow\] Area of \[ \Delta ABC = \dfrac{1}{2} \times 1 \times 2\left( {\sin {{60}^o}} \right)\]
By simplifying, we get
\[ \Rightarrow\] Area of \[ \Delta ABC = 1 \times \left( {\sin {{60}^o}} \right)\]
\[ \Rightarrow\] Area of \[\Delta ABC = 1 \times \left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow\] Area of \[\Delta ABC = \left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
Hence, the area of a triangle \[ABC\] is \[\dfrac{{\sqrt 3 }}{2}\].
Therefore, the correct option is (C).
Additional Information : The total space occupied by the three sides of every specific triangle is known as its area. The fundamental method of calculating the area of a triangle is half the product of its base and height. It is possible to find the area of a triangle in different ways. We can also find the area of a triangle in two different ways such as \[ A(\Delta ABC) = \dfrac{1}{2}bc\sin A\] and \[ A(\Delta ABC) = \dfrac{1}{2}ac\sin B\] .
Note:
Many students make mistakes in writing the formula of area of a triangle in terms of sine. This is the only way through which we can solve the example in the simplest way. Also, it is essential to simplify carefully and write the proper value of trigonometric angle to get the desired result.
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