In the triangle ABC AC = BC = 15, AB = 18. Find the sine of the outer corner at the vertex A.

The degree measure of the external angle A is equal to (180 ° – A). Sin (180 ° – A) = sinA.

Let’s draw the height of the CH. Since triangle ABC is isosceles, CH is also the median.

Hence, AH = 18: 2 = 9.

In a right-angled triangle ASN, according to the Pythagorean theorem:

CH² = AC² – AH² = 15² – 9² = 225 – 81 = 144.

CH = √144 = 12.

The sine of the angle is equal to the ratio of the opposite leg to the hypotenuse.

sinA = CH / AC = 12/15 = 4/5 = 0.8.

Answer: the sine of the outer angle at the vertex A is 0.8.