Question

to find the distance between a point x and an inaccessible point​ z, a line segment xy is constructed. measurements show that xy=966 ​m, angle xyz​=38°24', and angle yzx=94°6'. find the distance between x and z to the nearest meter.

Answer 1

Answer:

$\approx \bold{602\ m}$

Step-by-step explanation:

Given the following dimensions:

XY=966 ​m

$\angle XYZ$​ = 38°24', and

$\angle YZX$ = 94°6'

To find:

Distance between points X and Z.

Solution:

Let us plot the given values.

We can clearly see that it forms a triangle when we join the points X to Y, Y to Z and Z to X.

The $\triangle XYZ$ has following dimensions:

XY=966 ​m

$\angle XYZ$​ = 38°24', and

$\angle YZX$ = 94°6'

in which we have to find the side XZ.

Kindly refer to the image attached.

Let us use the Sine rule here:

As per Sine Rule:

$\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

Where  

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\dfrac{XZ}{sin\angle Y} = \dfrac{XY}{sin\angle Z}\\\Rightarrow \dfrac{966}{sin94^\circ6'} = \dfrac{XZ}{sin38^\circ24'}\\\Rightarrow XZ=\dfrac{966}{sin94^\circ6'} \times sin38^\circ24'\\\Rightarrow XZ=\dfrac{966}{0.997} \times 0.621\\\Rightarrow XZ=601.69 \m \approx \bold{602\ m}$