There are 3 sides to a Coin

Geometry Level pending

In an isosceles $\triangle ABC$ , $\angle A=100^\circ$ and $\angle B=\angle C=40^\circ$ . Side $AB$ is extended to point $D$ such that $AD=BC$ , where $D$ and $A$ are on the opposite sides of $B$ .

Find the measure of $\angle BCD$ in degrees.

Source-Adventures In Problem Solving by Shailesh Shirali

10 20 15 35

1 solution

Sachetan Debray
Feb 23, 2021

Let Angle BCD be 40-x degrees. Then Angle ADB= x degrees.

Using sine law in triangle ABC, and letting AC=AB=a and BC=b,

$\frac{sin 40}{sin 100}=\frac{a}{b}$

and,

$\frac{sin x}{sin(40-x)}=\frac{b}/{b-a}$

using these 2 equations and the trigonometric identity for sinA-sinB,

$\frac{cos70}{sin100}=\frac{sin(40-x)}{sin x}$

we can write cos 70 as sin 20 and sin 100 as sin80, we can then write sin 20 as sin 160 and further 2sin80cos80,

which means the LHS becomes sin 10 divided by sin 30(as 0.5 is sin 30 and cos 80 is sin(90-80),

so x=30.

Angle BCD=40-30=10 degrees.

There are 3 sides to a Coin

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