شرح حساب المثلثات في الفراغ: الأهرامات والمخاريط

Worked example

1. 1

   ${\text{base diagonal}}^2=8^2+8^2$ — The base is a square, so its diagonal is the longest side of a right triangle with two sides of 8. Square each side and add.

2. 2

   ${\text{base diagonal}}^2=128$ — 64 + 64 = 128.

3. 3

   $XD=\frac{1}{2}\sqrt{128}=\sqrt{32}$ — The centre $X$ is halfway along the diagonal, so $XD$ is half of it. Halving a root: $\frac{1}{2}\sqrt{128}=\sqrt{128÷4}=\sqrt{32}$.

4. 4

   $A{X}^2+X{D}^2=A{D}^2$ — Triangle $AXD$ has its right angle at $X$, because $AX$ stands straight up. The slant edge $AD$ is the longest side.

5. 5

   $A{X}^2=9^2-32$ — We want a short side, so take the other square away from the hypotenuse squared. Also $(\sqrt{32}{)}^2=32$.

6. 6

   $A{X}^2=49$ — 81 − 32 = 49.

7. 7

   $AX=7\text{ cm}$ — Take the square root to undo the squaring. A height cannot be negative.

Worked example

1. 1

   $\text{triangle }AXD\text{, right angle at }X$ — The slant edge $AD$ meets the base at the corner $D$. Its shadow on the base is $XD$, so the angle $\theta$ is at $D$.

2. 2

   $\tan \theta =\frac{AX}{XD}=\frac{7}{\sqrt{32}}$ — $AX$ is across from $\theta$ (opposite) and the half diagonal is next to it (adjacent). Opposite and adjacent mean tan.

3. 3

   $\theta ={\tan }^{-1}(1.2374\dots )$ — We know the ratio and want the angle. Inverse tan undoes tan.

4. 4

   $\theta =51.1^{\circ }$ — Round the angle to 1 decimal place.

Worked example

1. 1

   $XM=\frac{1}{2}\times 8=4$ — $M$ is the middle of a base edge. From the centre to $M$ is half the side.

2. 2

   $\text{triangle }AXM\text{, right angle at }X$ — The face slopes up from the edge at $M$, so the angle $\varphi$ between the face and the base is at $M$.

3. 3

   $\tan \varphi =\frac{7}{4}$ — $AX$ is opposite $\varphi$ and $XM$ is adjacent. Opposite and adjacent mean tan.

4. 4

   $\varphi =60.3^{\circ }$ — The face is steeper than the edge (51.1°) because $M$ is closer to the centre than the corner $D$ is.

Worked example

1. 1

   $d^2=12^2+6^2$ — The diagonal of the base is the longest side of a right triangle with sides 12 and 6.

2. 2

   $d^2=144+36=180$ — Square each side: 12 × 12 = 144 and 6 × 6 = 36. Then add.

3. 3

   $X{D}^2=180÷4=45$ — $XD$ is half the diagonal, so its square is a quarter of $d^2$.

4. 4

   $h^2=11^2-45$ — Right angle at $X$. The slant edge is the longest side, so subtract.

5. 5

   $h^2=76$ — 121 − 45 = 76.

6. 6

   $h=8.72\text{ cm}$ — Take the square root and round to 3 significant figures.

7. 7

   $\cos \theta =\frac{\sqrt{45}}{11}$ — The angle $\theta$ is at the corner. $XD=\sqrt{45}$ is next to it (adjacent) and the slant edge 11 is the hypotenuse. Adjacent and hypotenuse mean cos.

8. 8

   $\theta =52.4^{\circ }$ — Inverse cos on the calculator, rounded to 1 decimal place.

Worked example

1. 1

   $XM=\frac{1}{2}\times 10=5$ — $M$ is the middle of a base edge. From the centre $X$ to $M$ is half the side.

2. 2

   $l^2=12^2+5^2$ — The slant height $l$ is the longest side of the right triangle $AXM$, with the right angle at $X$.

3. 3

   $l^2=144+25=169$ — Square each side and add.

4. 4

   $l=13\text{ cm}$ — Take the square root.

5. 5

   $\text{area}=\frac{1}{2}\times 10\times 13$ — A triangle's area is half base × perpendicular height. The slant height is at right angles to the base edge, so it is the height of the face.

6. 6

   $\text{area}=65{\text{ cm}}^2$ — Multiply it out and add the unit.

Worked example

1. 1

   $\frac{1}{2}\times 8\times l=52$ — The area of the face is half base × slant height.

2. 2

   $4l=52,l=13$ — Half of 8 is 4. Divide both sides by 4.

3. 3

   $h^2=13^2-4^2$ — Right angle at the centre. The slant height is the longest side, and the centre is half the side, 4, from the middle of the edge.

4. 4

   $h^2=169-16=153$ — 13 × 13 = 169 and 4 × 4 = 16. Subtract.

5. 5

   $h=12.4\text{ cm}$ — Take the square root and round to 3 significant figures.

Worked example

1. 1

   $r=5,l=13$ — A cone has a radius $r$ and a slant height $l$. The slant height runs from the rim up to the tip.

2. 2

   $h^2=l^2-r^2$ — The right angle is at the centre of the base. The slant height is the longest side, so subtract.

3. 3

   $h^2=13^2-5^2=144$ — 169 − 25 = 144.

4. 4

   $h=12$ — Take the square root. $h$ is the vertical height of the cone.

5. 5

   $\cos \theta =\frac{5}{13}$ — $\theta$ is the angle between the slant and the base, at the rim. The radius is next to it (adjacent) and the slant is the hypotenuse. Adjacent and hypotenuse mean cos.

6. 6

   $\theta =67.4^{\circ }$ — Inverse cos, rounded to 1 decimal place.