Every card in Geometry and measures, Year 11: frustums and trigonometry

108π cm³

HintFind the two parts separately, remembering that only half a ball is used.

WhyCylinder: π × 3² × 10 = 90π. Whole sphere: 4/3 × π × 3³ = 36π, so the hemisphere is 18π. Total: 90π + 18π = 108π cm³.

144; 18; 126

HintWhat is left is a big cone with a little cone taken away.

WhyWhole cone: 1/3 × π × 6² × 12 = 144π. Removed cone: 1/3 × π × 3² × 6 = 18π. What is left (a frustum): 144π − 18π = 126π cm³. Although the cut is half-way up, the piece removed is only one eighth of the volume.

A frustum

HintThink of the shape of a bucket or a lampshade; the name begins with f.

WhyA frustum has two circular faces of different sizes joined by a sloping curved surface. Its volume is found by subtracting the small cone that was removed from the original cone.

54π cm²

HintList only the surfaces you could touch from outside: one flat circle and two curved surfaces.

WhyCurved surface of the cylinder: 2 × π × 3 × 5 = 30π. Its base: π × 3² = 9π. Curved surface of the cone: π × 3 × 5 = 15π. Total: 30π + 9π + 15π = 54π cm². The two circles where the shapes join are inside the solid and are not counted.

(200 − 10π) cm³

HintWork out the solid block first, then the shape of the material that was removed.

WhyCuboid: 4 × 5 × 10 = 200 cm³. The hole is a cylinder 10 cm long: π × 1² × 10 = 10π cm³. Remaining: 200 − 10π cm³, which is about 168.6 cm³.

inverse

HintIt is the function that undoes sine, in the way that subtracting undoes adding.

WhyIf sin x = 0.5 then x = sin⁻¹(0.5) = 30°. The same idea gives cos⁻¹ and tan⁻¹. On most calculators these are reached with the SHIFT key.

Cosine: A = cos⁻¹(5 ÷ 13)

HintLabel the two sides you know from A's point of view: is AB touching that corner or facing it?

WhyFrom angle A, AB is the adjacent side and AC is the hypotenuse, and the ratio that links those two is cosine. cos A = 5/13, so A = cos⁻¹(5/13), which is 67.4° to one decimal place.

30

HintWork out opposite ÷ hypotenuse and match it to an exact value you have learned.

Whysin x = 5/10 = 1/2. The angle whose sine is exactly 1/2 is 30°, one of the exact values to know.

14.0°

HintThe two lengths you know are the one facing the angle and the one beside it along the ground; neither is the sloping length.

WhyThe rise is opposite the angle and the horizontal distance is adjacent, so tan x = 1.5 ÷ 6 = 0.25. Then x = tan⁻¹(0.25) = 14.04°, which is 14.0° to one decimal place.

The fraction is upside down: cos x = adjacent ÷ hypotenuse = 5 ÷ 12, and a cosine can never be more than 1

HintThe longest side of the triangle belongs on the bottom here. How big can the result be?

WhySine and cosine both divide a shorter side by the hypotenuse, so for an angle in a right-angled triangle their values lie between 0 and 1. An error from sin⁻¹ or cos⁻¹ is a sign that the division was done the wrong way round. Here x = cos⁻¹(5/12) = 65.4° to one decimal place.

13

HintUse Pythagoras' theorem twice: first across one face, then from that diagonal to the far corner.

WhyThe diagonal of the 3 by 4 face is √(3² + 4²) = 5 cm. That diagonal and the 12 cm edge form a right-angled triangle whose hypotenuse is the diagonal through the cuboid: √(5² + 12²) = 13 cm. In one step, 3² + 4² + 12² = 169 = 13².

45°

HintFirst find the length of AC along the floor, then look at the triangle standing on it.

WhyAC is the diagonal of the base: √(4² + 3²) = 5 cm. Triangle ACG has a right angle at C, with AC = 5 cm and CG = 5 cm, so tan(angle GAC) = 5 ÷ 5 = 1 and the angle is 45°.

3; 5

HintO is in the middle of the square, so OM reaches only part of the way across. Then VOM is a right-angled triangle.

WhyFrom the centre of a square to the middle of a side is half the side length: 3 cm. VO is vertical and OM is horizontal, so triangle VOM has a right angle at O, and VM = √(4² + 3²) = 5 cm.

19.4°

HintThe observer's ground distance from the pole runs corner to corner across the field.

WhyThe diagonal of the field is √(8² + 15²) = 17 m. The pole, the diagonal and the line of sight form a right-angled triangle, so tan x = 6 ÷ 17 and x = tan⁻¹(6 ÷ 17) = 19.4° to one decimal place.

VMO (also written OMV)

HintYou need two lines that each meet AB squarely, one lying in each of the two surfaces.

WhyThe face and the base meet along AB. The angle between two planes is measured between lines, one in each plane, that are perpendicular to that shared edge. MV (in the face) and MO (in the base) are both perpendicular to AB at M, so the angle is VMO, found from tan(VMO) = VO ÷ OM.

1/2

HintPicture an equilateral triangle with sides of 2, cut in two from top to bottom.

WhyHalving an equilateral triangle of side 2 gives a right-angled triangle with angles of 30°, 60° and 90°. Opposite the 30° angle is half a side, 1, and the hypotenuse is 2, so sin 30° = 1/2.

√3/2 (root 3, divided by 2)

HintTake an equilateral triangle of side 2 and cut it down the middle. Find its height with Pythagoras' theorem; the height lies next to the 30° angle.

WhyThe cut triangle has hypotenuse 2 and shortest side 1, so its height is √(2² − 1²) = √3. The height is adjacent to the 30° angle, so cos 30° = √3/2, about 0.866.

√2/2 (the same as 1/√2); cos 45° has the same value

HintUse a right-angled triangle whose two shorter sides are both 1, and find its longest side.

WhyA right-angled triangle with two sides of 1 has angles of 45°, 45° and 90° and hypotenuse √2. So sin 45° = cos 45° = 1/√2, which is √2/2 when the denominator is rationalised, about 0.707. Because opposite and adjacent are equal, tan 45° = 1.

1

HintAt 90° the side facing the angle has become as long as the hypotenuse.

Whysin 0° = 0 and sin 90° = 1. Cosine runs the other way: cos 0° = 1 and cos 90° = 0. Also tan 0° = 0; the specification asks for no exact value of tan 90°.

5√3 cm

HintThe two sides involved are opposite and adjacent, so choose the ratio that links them and recall its exact value at 60°.

Whytan 60° = opposite ÷ adjacent, and tan 60° = √3 exactly. So opposite = 5 × √3 = 5√3 cm, about 8.66 cm. The other exact tangents are tan 30° = 1/√3 (or √3/3) and tan 45° = 1.