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Trigonometric Ratios and Identities · Paper 1 — Free Maths AI HL Practice Questions

1FoundationSAQ-SDefinitions of sine, cosine and tangent using right-angled triangles5 marksPaper 1~8 min
A ladder leans against a vertical wall. The base of the ladder is 1.5 m1.5\,\text{m} from the wall, and the ladder makes angle of 72°72° with the horizontal ground.
(a)
Show that the height, hh, the ladder reaches up the wall satisfies h=1.5tan⁡72°h = 1.5\tan 72°. [2 marks]
(b)
Calculate the length of the ladder. Give your answer correct to 3 significant figures. [2 marks]
(c)
The base of the ladder is pushed 0.3 m0.3\,\text{m} further from the wall. Determine the new angle the ladder makes with the ground, correct to 3 significant figures. [1 mark]
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Solutions

2MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles5 marksPaper 1~8 min
A surveyor measures the height of a vertical cliff. From point AA on level ground, the angle of elevation to the top of the cliff is 35∘35^\circ. The surveyor walks 50 m50\,\text{m} directly towards the cliff to point BB, from which the angle of elevation to the top of the cliff is 55∘55^\circ. Let hh metres be the height of the cliff and xx metres be the horizontal distance from BB to the base of the cliff.
(a)
Show that h=xtan⁡55∘h = x\tan 55^\circ. [1 mark]
(b)
Show that h=(x+50)tan⁡35∘h = (x + 50)\tan 35^\circ. [1 mark]
(c)
Find the value of xx, giving your answer correct to 3 significant figures. [2 marks]
(d)
Hence find the height of the cliff, giving your answer correct to 3 significant figures. [1 mark]
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Solutions

3ChallengeSAQ-LDefinitions of sine, cosine and tangent using right-angled triangles8 marksPaper 1~12 min
A surveyor measures the height of a vertical cliff face. From point AA on level ground, the angle of elevation to the top of the cliff CC is 35°35°. The surveyor walks 50 m50\,\text{m} directly towards the cliff to point BB. From BB, the angle of elevation to CC is 52°52°. Point DD is the base of the cliff. Let hh metres be the height of the cliff and xx metres be the horizontal distance BDBD.
(a)
Using triangle BCDBCD, write down an expression for hh in terms of xx. [1 mark]
(b)
Using triangle ACDACD and your result from part (a), show that x=50tan⁡35°tan⁡52°−tan⁡35°x = \frac{50\tan 35°}{\tan 52° - \tan 35°} [3 marks]
(c)
Hence calculate the height of the cliff hh, giving your answer correct to 3 significant figures. [2 marks]
(d)
The surveyor's measuring instrument is held at eye level, 1.6 m1.6\,\text{m} above the ground. Determine the percentage error introduced by ignoring this height when calculating hh. [2 marks]
diagram

Solutions

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