Question 13

We can find the third angle of a triangle using a line parallel to one of its sides.

Step 1 — Setting up the method

Let us draw a triangle. We will call this triangle ABC. Let angle B be one given angle. Let angle C be the other given angle. We need to find angle A, which is angle BAC.

Let us draw a line XY. This line passes through point A. Line XY is parallel to side BC.

Line AB is a transversal line. Since XY is parallel to BC, angle XAB equals angle ABC. These are called alternate interior angles. So, angle XAB equals angle B.

Line AC is also a transversal line. Since XY is parallel to BC, angle YAC equals angle ACB. These are also alternate interior angles. So, angle YAC equals angle C.

Line XAY is a straight line. Angles on a straight line add up to $180^\circ$. So, angle XAB + angle BAC + angle YAC = $180^\circ$. We will use this to find angle BAC.

Step 2 — Finding the third angle for $36^\circ, 72^\circ$

Let the first given angle be angle B. So, angle B is $36^\circ$. Let the second given angle be angle C. So, angle C is $72^\circ$. From Step 1, angle XAB equals angle B. So, angle XAB is $36^\circ$. From Step 1, angle YAC equals angle C. So, angle YAC is $72^\circ$. The sum of angles on line XY is $180^\circ$. So, angle XAB + angle BAC + angle YAC = $180^\circ$. Let us substitute the angle values.

$36^\circ + \text{angle BAC} + 72^\circ = 180^\circ$

$108^\circ + \text{angle BAC} = 180^\circ$

$\text{angle BAC} = 180^\circ - 108^\circ$

$\boxed{\text{angle BAC} = 72^\circ}$

Step 3 — Finding the third angle for $150^\circ, 15^\circ$

Let the first given angle be angle B. So, angle B is $150^\circ$. Let the second given angle be angle C. So, angle C is $15^\circ$. From Step 1, angle XAB equals angle B. So, angle XAB is $150^\circ$. From Step 1, angle YAC equals angle C. So, angle YAC is $15^\circ$. The sum of angles on line XY is $180^\circ$. So, angle XAB + angle BAC + angle YAC = $180^\circ$. Let us substitute the angle values.

$150^\circ + \text{angle BAC} + 15^\circ = 180^\circ$

$165^\circ + \text{angle BAC} = 180^\circ$

$\text{angle BAC} = 180^\circ - 165^\circ$

$\boxed{\text{angle BAC} = 15^\circ}$

Step 4 — Finding the third angle for $90^\circ, 30^\circ$

Let the first given angle be angle B. So, angle B is $90^\circ$. Let the second given angle be angle C. So, angle C is $30^\circ$. From Step 1, angle XAB equals angle B. So, angle XAB is $90^\circ$. From Step 1, angle YAC equals angle C. So, angle YAC is $30^\circ$. The sum of angles on line XY is $180^\circ$. So, angle XAB + angle BAC + angle YAC = $180^\circ$. Let us substitute the angle values.

$90^\circ + \text{angle BAC} + 30^\circ = 180^\circ$

$120^\circ + \text{angle BAC} = 180^\circ$

$\text{angle BAC} = 180^\circ - 120^\circ$

$\boxed{\text{angle BAC} = 60^\circ}$

Step 5 — Finding the third angle for $75^\circ, 45^\circ$

Let the first given angle be angle B. So, angle B is $75^\circ$. Let the second given angle be angle C. So, angle C is $45^\circ$. From Step 1, angle XAB equals angle B. So, angle XAB is $75^\circ$. From Step 1, angle YAC equals angle C. So, angle YAC is $45^\circ$. The sum of angles on line XY is $180^\circ$. So, angle XAB + angle BAC + angle YAC = $180^\circ$. Let us substitute the angle values.

$75^\circ + \text{angle BAC} + 45^\circ = 180^\circ$

$120^\circ + \text{angle BAC} = 180^\circ$

$\text{angle BAC} = 180^\circ - 120^\circ$

$\boxed{\text{angle BAC} = 60^\circ}$

Answer

(a) The third angle is $72^\circ$. (b) The third angle is $15^\circ$. (c) The third angle is $60^\circ$. (d) The third angle is $60^\circ$.