I was asked to calculate the area of the following isosceles triangle.

Let $\triangle ABC$ be an isosceles spherical triangle with $d(A,C)=d(B,C)$. Then $C$ lies on the perpendicular bisector of $AB$. Let $M$ be the midpoint of $AB$. Suppose $d(A,M)=d(-M,C)$. Calculate the area of $\triangle ABC$.

I already made a sketch and observed that $d(A,M)=d(-M,C)=\pi-d(M,C)$.

Let $\alpha$, $\beta$, $\gamma$ denote the interior angles corresponding to $A$, $B$, $C$ respectively.

If I assume without loss of generality that, because $C$ lies on the perpendicular bisector of $AB$, the angle $\gamma$ is divided into two equal angles $\gamma_1$ and $\gamma_2$ belonging to the right triangles $\triangle AMC$ and $\triangle BMC$. Since $$\operatorname{area}(\triangle ABC) = 2 \cdot \operatorname{area}(\triangle AMC) = 2 \cdot \operatorname{area}(\triangle BMC)$$ I only need to calculate $\operatorname{area}(\triangle AMC)$.

Since $\operatorname{area}(\triangle AMC)=\alpha + \gamma_1+ \frac{\pi}{2}-\pi$, I need $\alpha$ and $\gamma_1$. I first tried to calculate $\gamma_1$ using the spherical Law of Cosines, and I got this:

$$\cos\gamma_1=\cos\alpha \cdot \cos\frac{\pi}{2} - \sin\alpha \cdot \sin\frac{\pi}{2} \cdot \cos d(A,M)=-\sin\alpha \cdot \cos\left(\pi-d(M,C)\right) \tag{1}$$

From the spherical Law of Sines, I get: $$\sin\alpha= \frac{\sin d(M,C)}{\sin d(A,C)} \tag{2}$$

But now I do not see how to proceed. I would like to calculate the right-hand sides of $(1)$ and $(2)$, but this does not seem possible with the data I have. Or is this meant to be solved another way? I am not allowed to use Napier's rule since it was not proven in the lecture.


A more geometric approach starts with extending the line segments $MC$ and $MA$ to their meeting point $-M$.

This creates a bi-angle between $M$ and $-M$ with area $\pi.\, AC$ divides this bi-angle into two congruent triangles because $d(A,M) = d(C,-M)$, so $\bigtriangleup$$AMC$ has area $\frac\pi2$ and $\bigtriangleup$$ABC$ has area $\pi$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.