Find the Length of Inscribed Triangle using Trigonometry Can someone help me solve this question, I'm trying to teach my younger sister but was unable to find the correct answer. It's basic trigonometry, she has learnt the sine law, cosine law and unit circle trig ratios.
I thought the side angles had to be 45 degrees of the inscribed triangle and could then determine other angles starting with N then L, however that did not work. The answer according to the text is: 7.2m.

 A: Here is a simpler solution:

*

*Draw the perpendicular from $L$ to $MN$. Since the $\triangle MLN$ is isosceles, the height is also a median. Let the foot of the perpendicular from $L$ be called $F$.

*Calculate the height of the perpendicular $LF$ using Pythagoras' theorem. The hypotenuse is $LM=5.3$, one side is $MF=6.5/2$.

*In right angle triangle $LFK$ use the sine of angle $FKL$ to get the length of $KL$
A: Yet another solution, around the same lines as heropup, but slightly simpler.
Let $H$ be the projection of $L$ onto $MN$. A basic property of isosceles triangles give $HM=3.25$. Therefore, if we set $\alpha := \angle MNL$:
$$\cos \alpha = \frac{HN}{LM} = \frac{3.25}{5.3}$$
giving $\alpha \approx 52.18°$
Then sine law in triangle $KLM$ gives :
$$\frac{KL}{\sin 52.18°}=\frac{5.3}{\sin 35.3°}$$
giving $KL$.
A: Here is an outline of the solution process. The actual calculations are left as an exercise.

*

*Draw an altitude from $L$ to $NM$, creating two congruent right triangles from isosceles $\triangle NLM$.  Let the foot of the perpendicular from $L$ be called $F$.

*Using trigonometric ratios in either $\triangle NFL$ or $\triangle MFL$, compute the measure of $\angle NML = \angle MNL$.

*Use the fact that $\angle KNL + \angle MNL = 180^\circ$.

*Use the Law of Sines in $\triangle KNL$ to compute $KL$.

Alternatively, you may replace steps $1$ and $2$ above by using the Law of Cosines in $\triangle NML$.
