QuestionCalculate the distance $BC$ in triangle $ABC$ where $AB=1.5 \text{ km}$ , $AC=1.2 \text{ km}$ , and $\angle A=50^\circ$ .

Studdy Solution

STEP 1

Assumptions1. We have a triangle ABC. The length of AB is1.5 km3. The length of AC is1. km4. The angle at A is50 degrees5. We need to find the length of BC

STEP 2

We can use the Law of Cosines to find the length of BC. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and an angle γ opposite side c, the following equation holds $c^2 = a^2 + b^2 -2ab \cos(\gamma)$

STEP 3

In our case, side c corresponds to BC, side a corresponds to AB, side b corresponds to AC, and γ corresponds to the angle at A. So we can rewrite the Law of Cosines as $BC^2 = AB^2 + AC^2 -2AB \cdot AC \cos(A)$

STEP 4

Now, plug in the given values for AB, AC, and A to calculate BC.
$BC^2 = (1. \, km)^2 + (1.2 \, km)^2 -2 \cdot1. \, km \cdot1.2 \, km \cos(50^{\circ})$

STEP 5

Calculate the square of BC.
$BC^2 =2.25 \, km^2 +1.44 \, km^2 -2 \cdot1.5 \, km \cdot1.2 \, km \cdot \cos(50^{\circ})$

STEP 6

Calculate the cosine of50 degrees.
$\cos(50^{\circ}) \approx0.643$

STEP 7

Substitute the value of $\cos(50^{\circ})$ into the equation.
$BC^2 =2.25 \, km^2 +1.44 \, km^2 -2 \cdot1.5 \, km \cdot1.2 \, km \cdot0.643$

STEP 8

Calculate the right side of the equation.
$BC^2 \approx2.25 \, km^2 +1.44 \, km^2 -1.16 \, km^2$

STEP 9

Calculate the value of $BC^2$ .
$BC^2 \approx2.53 \, km^2$

STEP 10

To find BC, we need to take the square root of $BC^2$ .
$BC = \sqrt{BC^2}$

STEP 11

Plug in the value for BC^ to calculate BC.
BC = \sqrt{.53 \, km^}

STEP 12

Calculate the value of BC.
$BC \approx.59 \, km$ The distance BC is approximately.59 km.

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