Question

Circular tracts of land with diameters 900 meters, 700 meters and 600 meters are tangent to each other externally. There are houses directly in the center of each circle. What are the angles of the triangle connecting the houses and what is the area of that triangle?

Answer 1

Answer:

$\alpha=69.28^o$

$\beta=61.26^o$

$\gamma=49.46^o$

<em />$A=227980.26\ m^2$<em />

Step-by-step explanation:

Triangle Solving

If we had a triangle will its three sides of known length, we can solve for the rest of the parameters of the triangle, i.e. the area, perimeter and internal angles.

The three circles have diameters 900 m, 700 m and 600 m and are tangent to each other externally. The distances from their centers (where houses are located) are the sum of each pair of the radius of the circles. Thus, the sides of the triangle are

$x=450+350=800$

$y=450+300=750$

$z=350+300=650$

The internal angles can be computed by using the cosine's law

$x^2=y^2+z^2-2yzcos\alpha$

$y^2=x^2+z^2-2xzcos\beta$

$z^2=x^2+y^2-2xycos\gamma$

Where \alpha, \beta and \gamma are the opposite angles to x, y and z respectively. Solving for each one of them:

$\displaystyle cos\alpha=\frac{y^2+z^2-x^2}{2yz}$

$\displaystyle cos\alpha=\frac{750^2+650^2-800^2}{2\cdot 750\cdot 650}$

$cos\alpha=0.3538$

$\alpha=69.28^o$

Similarly

$\displaystyle cos\beta=\frac{x^2+z^2-y^2}{2xz}$

$\displaystyle cos\beta=\frac{800^2+650^2-750^2}{2\cdot 800\cdot 650}$

$cos\beta=0.4808$

$\beta=61.26^o$

The other angle is computed now

$\displaystyle cos\gamma=\frac{x^2+y^2-z^2}{2xy}$

$\displaystyle cos\gamma=\frac{800^2+750^2-650^2}{2\cdot 800\cdot 750}$

$cos\gamma=0.65$

$\gamma=49.46^o$

The area can be found by

$\displaystyle A=\frac{1}{2}x.y.sin\gamma$

$\displaystyle A=\frac{1}{2}\cdot 800\cdot 750\cdot sin49.46^o$

$A=227980.26\ m^2$