Question

In ΔTUV, u = 5.2 cm, ∠T=42° and ∠U=56°. Find the length of t, to the nearest 10th of a centimeter.

Answer 1

Answer

We have determined that any point on the pre-image figure is the same distance from the line of reflection as its image. Therefore, the two points are equidistant from the point at which the line of reflection (perpendicular bisector) intersects the segment connecting the pre-image point to its image.

Answer 2

According to the question,

In ΔTUV,

$u = 5.2 cm,$

∠$T=42^o$ and

∠$U=56^o$

As we have to find the length of t, nearest to the $10th$ of the cm.

So, $\frac{sin(T)}{t}=\frac{sin(U)}{u}$

Putting the given values,

$=>\frac{sin(42)}{t} =\frac{sin(56)}{5.2}$

By using cross multiplication,

$=>(t)sin(56)=(5.2)sin(42)$

Now, divide both sides by $sin (56),$

$=>\frac{(t)sin(56)}{sin(56)}=\frac{(5.2)sin(42)}{sin(56)}$

$=>t=\frac{(5.2)sin(42)}{sin(56)}$

$=>t=4.19701cm$

$=>t$ ≈ $4.2cm$

Hence, the length of t, nearest to the $10th$ of the cm is $4.2cm$.