Question

Given: ∆ABC, AB = BC, BM = MC AC = 40, m∠BAC = 42º Find: AM

Answer 1

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.

$\angle B=180-42-42=96$

Law of Sine

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$\frac{\sinB}{AC}=\frac{\sin(C}{AB}$

$\frac{\sin(96)}{40}=\frac{\sin(42}{AB}$

$AB\sin(96)=40\sin(42)$

$AB=\frac{40\sin(42)}{\sin(96)}$

$AB=26.91$

Therefore the length of AB and BC is 26.91.

Since M is midpoint of BC, so

$BM=\frac{BC}{2}=\frac{26.91}{2}=13.455$

Use Law of Cosine in triangle ABM to find the value of AM.

$a^2=b^2+c^2-2bc\cos A$

$AM^2=AB^2+BM^2-2(AB)(BM)\cos (B)$

$AM^2=(26.91)^2+(13.455)^2-2(26.91)(13.455)\cos (96)$

$AM=26.50$

Therefore the length of AM is 26.50 units.