In an isosceles triangle ABC (AB = BC), the angle B is 7 times the angle A. Find the outer angle BCD.

| April 6, 2021 | education

In order to find the value of the external angle BCD in an isosceles triangle ABC, if you know that angle B is 7 times the angle A, you first need to find out the value of angle C.
1.) Now we denote the value of the angle A through the variable x.
The angle B is then 7x.
Since ΔABC is isosceles, then, according to the rule, in this triangle, the angles at the base are equal.
A + B + C = 180 °.
Since ΔABC is isosceles, then, according to the rule, in this triangle, the angles at the base are equal.
That is, the angle A is equal to the angle C, so the angle C is also equal to x.
Now let’s find the value of the angle C.
x + 7x + x = 180 °.
9x = 180 °.
x = 180 °: 9.
x = 20 °.
2.) Now let’s find the value of the outer angle BCD.
To do this, recall one of the properties of the outer corners of the triangle.
By this property, the sum of the outer and inner angles of a triangle at one vertex is 180 °.
Now let’s write for the outer and inner angles ΔABC and find the angle BCD.
ACB + BCD = 180 °.
20 ° + BCD = 180 °.
BCD = 180 ° – 20 °.
BCD = 160 °.
Answer: 160 °.



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