# A triangle with angle A 15 degrees, and sides AB=AC=10cm, how do you find the all the angles and sides of the triangle?

Oct 22, 2015

$\angle B = \angle C = {82.5}^{o}$
$B C = 10 \sqrt{2 - \sqrt{2 + \sqrt{3}}} \approx 2.610524$

#### Explanation:

Draw a circle of a radius $A B = 10$ with a center $A$ and choose on it a point and label it $B$.
From point $B$, using the same radius $10$, draw an arc that intersects a circle at point $F$.
Obviously, $\Delta A B F$ is equilateral triangle, $A B = B F = A F = 10$ and $\angle B A F = {60}^{o}$.

Bisect $\angle B A F$ by a radius $A D$, so $\angle B A D = \angle D A F = \angle {30}^{o}$.
Bisect $\angle B A D$ by a radius $A C$, so $\angle B A C = \angle C A D = \angle {15}^{o}$.

We are ready now to derive the length $B C$ from $B D$, which, in turn, we can derive from $B F$, that we know is equal to 10.

Let $A D \cap B F = P$
Using Pythagorean Theorem, calculate $B D$ from $B P$ and $P D$.

$B {D}^{2} = B {P}^{2} + P {D}^{2}$
$B P = \frac{B F}{2} = \frac{10}{2} = 5$
$P D = A D - A P$
$A D = 10$
$A {P}^{2} = A {F}^{2} - P {F}^{2} = {10}^{2} - {\left(\frac{10}{2}\right)}^{2} = 100 - 25 = 75$
$A P = \sqrt{75} = 5 \sqrt{3}$
$P D = 10 - 5 \sqrt{3}$
$B {D}^{2} = {\left(\frac{10}{2}\right)}^{2} + {\left(10 - 5 \sqrt{3}\right)}^{2} = 25 + 100 - 100 \sqrt{3} + 75 =$
$= 100 \left(2 - \sqrt{3}\right)$
$B D = 10 \sqrt{2 - \sqrt{3}} \approx 5.176381$

Analogously, calculate $B C$ by knowing $B D$.
Let $A C \cap B D = Q$
Using Pythagorean Theorem, calculate $B C$ from $B Q$ and $Q C$.

$B Q = \frac{B D}{2} = 5 \sqrt{2 - \sqrt{3}}$
$Q C = A C - A Q$
$A C = 10$
$A {Q}^{2} = A {B}^{2} - B {Q}^{2} = {10}^{2} - 25 \left(2 - \sqrt{3}\right) = 25 \left(2 + \sqrt{3}\right)$
$A Q = 5 \sqrt{2 + \sqrt{3}}$
$Q C = 10 - 5 \sqrt{2 + \sqrt{3}}$
$B {C}^{2} = B {Q}^{2} + Q {C}^{2} =$
$= 25 \left(2 - \sqrt{3}\right) + 100 - 100 \sqrt{2 + \sqrt{3}} + 25 \left(2 + \sqrt{3}\right) =$
$= 100 \left(2 - \sqrt{2 + \sqrt{3}}\right)$
$B C = 10 \sqrt{2 - \sqrt{2 + \sqrt{3}}} \approx 2.610524$