Answer:
100°
Explanation:
You want the measure of angle θ = ∠BDC in the figure of isosceles triangle ABC with vertex angle A=40° and interior point D such that AD≅BC and ∠BAD = 10°.
Vector solution
We can define the midpoint of BC as the origin of a coordinate plane, and points B(-1, 0) and C(1, 0). Then point A will be located at A(0, tan(70°)).
Point D will be located at 2∠-100° from A, since AD=BC=2 units, and D is located 10° west of south relative to A.
This locates point D on the coordinate grid.
To find the angle between DC and DB, we can divide vector DC by vector DB. The angle of the quotient is the angle between those vectors. The calculator result in the second attachment shows the angle is 100°, in agreement with the geometry program's result in the first attachment.
θ = 100°
Law of cosines solution
Given our assignment of a length of 2 to AD and BC, we know the lengths of AC and AB are sec(70°) ≈ 2.9238. This lets us use the law of cosines to find lengths DC and DB:
CD = √(2.9238² +2² -2·2.9238·2·cos(30°)) ≈ 1.5557
DB = √(2.9238² +2² -2·2.9238·2·cos(10°)) ≈ 1.0154
Then the law of cosines can be used again to find θ.
θ = arccos((1.5557² +1.0154² -2²)/(2·1.5557·1.1054))
θ = 100°
Geometrical solution
The third attachment shows a geometrical solution to finding the angle. We define O as the center of the circle. Angle BAC is an inscribed angle of 40°, so angle BOC will be double that value, 80°.
Locating point G so that AOG is 80° makes segment AG the same length as segments AB and AD. AO bisects angle A, so angle GAD is 10° less than angle GAO, making ∆GAD~∆ACG and locating point D on segment AG. While point D can be shown to lie on segment AG using angle sums, it takes Law of Sines and Law of Cosines relations to demonstrate that point D lies on segment BO, which it does.
The angle of interest, BDC, is an external angle to triangle COD. As such, its measure is the sum of the remote interior angles of that triangle:
∠BDC = 80° +20°
∠BDC = 100°