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

Studdy Solution

STEP 1

Assumptions1. Triangle ABC is a non-right triangle. The lengths of AB and AC are given as1.5 km and1. km respectively3. The angle A is given as50 degrees

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, we are trying to find the length of BC, which we can denote as c. The lengths of AB and AC are a and b respectively, and the angle A is γ. So we can write the equation as $c^2 = a^2 + b^2 -2ab \cos(A)$

STEP 4

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

STEP 5

First, square the lengths of AB and AC.
$c^2 =2.25 \, km^2 +1.44 \, km^2 -2(1.5 \, km)(1.2 \, km) \cos(50^{\circ})$

STEP 6

Next, calculate the cosine of the angle A.
$\cos(50^{\circ}) \approx0.6428$

STEP 7

Then, plug in the value of the cosine of A into the equation.
$c^2 =2.25 \, km^2 +1.44 \, km^2 -2(1.5 \, km)(1.2 \, km) \times0.642$

STEP 8

Perform the multiplication and subtraction in the equation.
$c^2 \approx2.25 \, km^2 +1.44 \, km^2 -1.16 \, km^2$

STEP 9

Add the squares of the lengths of AB and AC, and subtract the result of the multiplication.
$c^2 \approx2.53 \, km^2$

STEP 10

Finally, take the square root of both sides of the equation to solve for c.
$c \approx \sqrt{2.53 \, km^2}$

STEP 11

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

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