Question

4. How many triangles are created in the following situation:
ZA= 51°, a = 22,b=27
Notebo...
O i triangle
O 3 triangles
O no triangle
O 2 triangles

Answer 1

Only one triangle can be created for $A = 51^{\circ}$, $a = 22$ and $b = 27$.

Determination of the number of possible triangles based on known sides and angles

A triangle is a geometrical figure with three sides and three internal angles, whose sum equals 180°. Let be the triangle ABC, where $A = 51^{\circ}$, $a = 22$ and $b = 27$, where $A$ is the angle opposite to the side $a$.

By law of cosine we find that the two possible measures of the side $c$ are:

$a^{2} = b^{2} + c^{2}-2\cdot b\cdot c \cdot \cos A$ (1)

$c^{2} - 33.983\cdot c +245 = 0$

$c_{1} \approx 23.603$, $c_{2}\approx 10.380$

And by law of sine we have the measures of angles $B$ and $C$ for each case:

(i) $c_{1} \approx 23.603$

$B = \sin^{-1}\left[\left(\frac{b}{a} \right)\cdot \sin A\right]$

$B \approx 72.510^{\circ}$

$C = \sin^{-1}\left[\left(\frac{c}{a} \right)\cdot \sin A\right]$

$C \approx 56.488^{\circ}$

(ii) $c_{2}\approx 10.380$

$B = \sin^{-1}\left[\left(\frac{b}{a} \right)\cdot \sin A\right]$

$B \approx 72.510^{\circ}$

$C = \sin^{-1}\left[\left(\frac{c}{a} \right)\cdot \sin A\right]$

$C \approx 21.510^{\circ}$

By geometry we find that the only realistic solution is found by $c_{1} \approx 23.603$, as it satisfies the condition of the sum of internal angles. Hence, only one triangle can be created for $A = 51^{\circ}$, $a = 22$ and $b = 27$. $\blacksquare$

To learn more on triangles, we kindly invite to check this verified question: https://brainly.com/question/24005275