Question

question 7
below is a triangular pyramid box a candy store uses on special requests for themed birthday parties. it consists of four triangles. line $ab = \sqrt{3}x$, $cb = x$, $a\hat{c}d=alpha$, and $a\hat{d}b = 60^{circ}$. ab is perpendicular to cb and bd.
nsc sept: mathematics p2 | 2025
7.2 prove that $a\hat{c}d = a\hat{d}c$.
(3)
7.3 prove that $cd = 4x\cos\alpha$.
(4)
9

Explanation:

Step1: Find length of AD in right - triangle ABD

In right - triangle ABD, $\angle ADB = 60^{\circ}$ and $AB=\sqrt{3}x$. Since $\tan\angle ADB=\frac{AB}{BD}$ and $\sin\angle ADB=\frac{AB}{AD}$, and $\sin60^{\circ}=\frac{\sqrt{3}}{2}=\frac{AB}{AD}$, substituting $AB = \sqrt{3}x$ gives $AD = 2x$.

Step2: Find length of AC in right - triangle ABC

In right - triangle ABC, $AB=\sqrt{3}x$ and $CB = x$. By the Pythagorean theorem $AC=\sqrt{AB^{2}+CB^{2}}=\sqrt{(\sqrt{3}x)^{2}+x^{2}}=\sqrt{3x^{2}+x^{2}} = 2x$.

Step3: Prove $\angle ACD=\angle ADC$

Since $AC = AD=2x$ in $\triangle ACD$, by the property of an isosceles triangle (sides opposite equal angles are equal), we have $\angle ACD=\angle ADC$.

Step4: Prove $CD = 4x\cos\alpha$

In $\triangle ACD$, using the cosine - law $CD^{2}=AC^{2}+AD^{2}-2\cdot AC\cdot AD\cdot\cos\angle CAD$. Since $AC = AD = 2x$, we have $CD^{2}=4x^{2}+4x^{2}-2\times2x\times2x\times\cos(180^{\circ}-2\alpha)$. Using the double - angle formula $\cos(180^{\circ}-2\alpha)=-\cos2\alpha=-(2\cos^{2}\alpha - 1)=1 - 2\cos^{2}\alpha$. Then $CD^{2}=8x^{2}-8x^{2}(1 - 2\cos^{2}\alpha)=16x^{2}\cos^{2}\alpha$. Taking the square root of both sides (since $CD>0$), we get $CD = 4x\cos\alpha$.

Answer:

7.2: Since $AC = AD = 2x$ in $\triangle ACD$, $\angle ACD=\angle ADC$ (sides opposite equal angles in an isosceles triangle are equal).
7.3: In $\triangle ACD$, by the cosine - law and simplification using trigonometric identities, $CD = 4x\cos\alpha$.