
In $\vartriangle {\text{ABC}}$, $2\angle {\text{A = 3}}\angle {\text{B}}$, $5\angle {\text{B = 2}}\angle {\text{C}}$. Then the triangle is
A. Isosceles
B. Right – angled
C. Equilateral
D. Right – angled and isosceles
Answer
600.6k+ views
Hint: In order to solve this question, we will apply the angle – sum property of a triangle because the relation between the angles of the triangle is given.
Complete step-by-step answer:
We are given a relation between angles of the triangle. Here we will use the angle sum property of a triangle. This property states that the sum of all three angles is equal to ${180^0}$. Therefore,
$\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C = 18}}{{\text{0}}^0}$ …….. (1)
As, $2\angle {\text{A = 3}}\angle {\text{B}}$ and $5\angle {\text{B = 2}}\angle {\text{C}}$. Putting these values in equation (1), we get
$ \Rightarrow $ $\dfrac{{3\angle {\text{B}}}}{2}{\text{ + }}\angle {\text{B + }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = 18}}{{\text{0}}^0}$
$ \Rightarrow $ $\angle {\text{C = }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{5 \times {{36}^0}}}{2}{\text{ = 9}}{{\text{0}}^0}$
$ \Rightarrow $ $\angle {\text{B = 3}}{{\text{6}}^0}$
Now applying the value of $\angle {\text{B }}$in the given conditions. So,
$\angle {\text{A = }}\dfrac{{3\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{3 \times {{36}^0}}}{2}{\text{ = 5}}{{\text{4}}^0}$
$\angle {\text{C = }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{5 \times {{36}^0}}}{2}{\text{ = 9}}{{\text{0}}^0}$
So, the triangle looks like the figure as shown below.
As, we can see from the figure that the triangle has a right angle. So, it is a right – angled triangle. So, option (B) is the answer.
Note: In such types of questions after applying the angle sum property and after finding all three angles double check your answer by adding all the angles so that their sum is equal to ${180^0}$.Apply proper values in the condition to find the correct answer. After finding the angles you should also check all the options given to find the correct answer. For example, if a triangle by finding angles turns out to be right angle but it can also be an isosceles triangle which you can find if you make a proper figure and check all the options given.
Complete step-by-step answer:
We are given a relation between angles of the triangle. Here we will use the angle sum property of a triangle. This property states that the sum of all three angles is equal to ${180^0}$. Therefore,
$\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C = 18}}{{\text{0}}^0}$ …….. (1)
As, $2\angle {\text{A = 3}}\angle {\text{B}}$ and $5\angle {\text{B = 2}}\angle {\text{C}}$. Putting these values in equation (1), we get
$ \Rightarrow $ $\dfrac{{3\angle {\text{B}}}}{2}{\text{ + }}\angle {\text{B + }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = 18}}{{\text{0}}^0}$
$ \Rightarrow $ $\angle {\text{C = }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{5 \times {{36}^0}}}{2}{\text{ = 9}}{{\text{0}}^0}$
$ \Rightarrow $ $\angle {\text{B = 3}}{{\text{6}}^0}$
Now applying the value of $\angle {\text{B }}$in the given conditions. So,
$\angle {\text{A = }}\dfrac{{3\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{3 \times {{36}^0}}}{2}{\text{ = 5}}{{\text{4}}^0}$
$\angle {\text{C = }}\dfrac{{5\angle {\text{B}}}}{2}{\text{ = }}\dfrac{{5 \times {{36}^0}}}{2}{\text{ = 9}}{{\text{0}}^0}$
So, the triangle looks like the figure as shown below.
As, we can see from the figure that the triangle has a right angle. So, it is a right – angled triangle. So, option (B) is the answer.
Note: In such types of questions after applying the angle sum property and after finding all three angles double check your answer by adding all the angles so that their sum is equal to ${180^0}$.Apply proper values in the condition to find the correct answer. After finding the angles you should also check all the options given to find the correct answer. For example, if a triangle by finding angles turns out to be right angle but it can also be an isosceles triangle which you can find if you make a proper figure and check all the options given.
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