In the diagram below, B1=beta, B2=alpha, AB=y units, BD=DC=x units and AC BC. (1) Show that D=180⁻²alpha. (2) Prove that y=(x 2alpha)/( alpha beta). (3) If x=150 m, alpha=30^ and beta=70^, calculate y.

In $\triangle BCD$, we are given that $BD = DC = x$ units. This means $\triangle BCD$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal. We are given $\hat{B}_2 = \alpha$, which is $\angle DBC$. Therefore, $\angle DBC = \angle DCB = \alpha$.

The sum of angles in any triangle is $180^\circ$. For $\triangle BCD$:

Substitute the known angles:

Rearrange to solve for $\hat{D}$:

This proves the statement.

In $\triangle BCD$, we have sides $BD=x$, $DC=x$, and angles $\angle DBC=\alpha$, $\angle DCB=\alpha$, and $\hat{D}=180^\circ - 2\alpha$.
Applying the Sine Rule:

Substitute the known values:

Using the identity $\sin(180^\circ - \theta) = \sin \theta$:

Solve for $BC$:

We are given that $AC \perp BC$, so $\triangle ABC$ is a right-angled triangle at C.
We have $\hat{B}_1 = \beta$, which is $\angle ABC$. The hypotenuse is $AB = y$.
In a right-angled triangle, $\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}$.

Substitute $AB = y$:

Solve for $y$:

Substitute $BC = \frac{x \sin 2\alpha}{\sin \alpha}$ into $y = \frac{BC}{\cos \beta}$:

This proves the statement.

Given $x = 150$ m, $\alpha = 30^\circ$, $\beta = 70^\circ$.

Substitute these values:

Using a calculator:

Rounding to two decimal places:

The value of $y$ is $\boxed{\text{759.64 m}}$.