Question

Cosine Rule In ABC, BC = 4cm, M is the mid-point of BC, AM = 4cm and AMB = 120..... Check the image below

Answer 1

Answer:

$\sf (a) \quad AC=2\sqrt{3}\:\:cm$

$\sf (b) \quad AB=2\sqrt{7}\:\:cm$

$\sf (c) \quad \angle ACB=90^{\circ}$

Step-by-step explanation:

Cosine rule

$\sf c^2=a^2+b^2-2ab \cos C$

where:

• a, b and c are the sides of the triangle.
• C is the angle opposite side c.

Sketch the triangle using the given information (see attached).

Part (a)

Given:

• a = MC = 2
• b = AM = 4
• c = AC
• C = ∠AMC = 60°

Substitute the given values into the formula and solve for AC:

$\implies \sf c^2=a^2+b^2-2ab \cos C$

$\implies \sf AC^2=2^2+4^2-2(2)(4) \cos 60^{\circ}$

$\implies \sf AC^2=4+16-16 \left(\dfrac{1}{2}\right)$

$\implies \sf AC^2=20-8$

$\implies \sf AC^2=12$

$\implies \sf AC=\sqrt{12}$

$\implies \sf AC=\sqrt{4 \cdot 3}$

$\implies \sf AC=\sqrt{4}\sqrt{3}$

$\implies \sf AC=2\sqrt{3}\:\:cm$

Part (b)

Given:

• a = BM = 2
• b = AM = 4
• c = AB
• C = ∠AMB = 120°

Substitute the given values into the formula and solve for AB:

$\implies \sf c^2=a^2+b^2-2ab \cos C$

$\implies \sf AB^2=2^2+4^2-2(2)(4) \cos 120^{\circ}$

$\implies \sf AB^2=4+16-16 \left(-\dfrac{1}{2}\right)$

$\implies \sf AB^2=20+8$

$\implies \sf AB^2=28$

$\implies \sf AB=\sqrt{28}$

$\implies \sf AB=\sqrt{4\cdot7}$

$\implies \sf AB=\sqrt{4}\sqrt{7}$

$\implies \sf AB=2\sqrt{7}\:\:cm$

Part (c)

Given:

• a = AC = 2√3
• b = BC = 4
• c = AB = 2√7
• C = ∠ACB

Substitute the given values into the formula and solve for ∠ACB:

$\implies \sf c^2=a^2+b^2-2ab \cos C$

$\implies \sf \left(2\sqrt{7}\right)^2=\left(2\sqrt{3}\right)^2+4^2-2\left(2\sqrt{3}\right)(4) \cos ACB$

$\implies \sf 28=12+16-16\sqrt{3} \cos ACB$

$\implies \sf 28=28-16\sqrt{3} \cos ACB$

$\implies \sf 0=-16\sqrt{3} \cos ACB$

$\implies \sf \cos ACB=0$

$\implies \sf ACB=\cos^{-1}(0)$

$\implies \sf ACB=90^{\circ}$