Answer:
The length of AM is 26.50 units.
Step-by-step explanation:
Given information: AB = BC, BM = MC , AC = 40, ∠BAC = 42º.
Since two sides of triangle are equal, therefore the triangle ABC is an isosceles triangle.
The corresponding angles of congruents sides are always equal. So angle C is 42º.
According to the angle sum property the sum of interior angles is 180º.
[tex]\angle B=180-42-42=96[/tex]
Law of Sine
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]
[tex]\frac{\sinB}{AC}=\frac{\sin(C}{AB}[/tex]
[tex]\frac{\sin(96)}{40}=\frac{\sin(42}{AB}[/tex]
[tex]AB\sin(96)=40\sin(42)[/tex]
[tex]AB=\frac{40\sin(42)}{\sin(96)}[/tex]
[tex]AB=26.91[/tex]
Therefore the length of AB and BC is 26.91.
Since M is midpoint of BC, so
[tex]BM=\frac{BC}{2}=\frac{26.91}{2}=13.455[/tex]
Use Law of Cosine in triangle ABM to find the value of AM.
[tex]a^2=b^2+c^2-2bc\cos A[/tex]
[tex]AM^2=AB^2+BM^2-2(AB)(BM)\cos (B)[/tex]
[tex]AM^2=(26.91)^2+(13.455)^2-2(26.91)(13.455)\cos (96)[/tex]
[tex]AM=26.50[/tex]
Therefore the length of AM is 26.50 units.
Answer:
31.32208078, for those that need the more accurate answer. The steps are pretty much the same as the first answer. Just thought this would be helpful.
