StudentAidUnit 7: Trigonometry — Exercise 7 3
10th Class Mathematics · Unit 7: Trigonometry
7.3.1.In the triangular prism (AB=3 cm, FE=5 cm, EC=4 cm), find (i) the length CF (ii) the length BF (iii) the angle BFC, correct to one decimal place.
Formula
(i) Pythagoras in right ΔCEF
CF2=CE2+EF2=42+52=41
Result
(i)
CF=sqrt{41}=6.40 cm
Formula
(ii) Pythagoras in right ΔBCF
BF2=BC2+CF2=32+41=50
Result
(ii)
BF=sqrt{50}=5sqrt2=7.07 cm
Formula
(iii) Cosine Rule in ΔBCF
BC2=BF2+CF2-2(BF)(CF)cosangle BFC
9=91-2sqrt{2050}cosθ ⇒ cosθ=frac{41}{sqrt{2050}=0.9050
Result
(iii)
angle BFC=cos-1(0.9050)=25.0^circ
7.3.2.The diagram shows a triangular pyramid with a horizontal rectangular base PQRS, in which m PQ = 16 cm, m QR = 10 cm. M is the midpoint of the line PR. The vertex, T, is vertically above M and m MT = 15 cm. Calculate the size of the angle between TP and the base PQRS. Give your answer correct to 1 decimal place.
Given
The angle between TP and the base is angle TPM
Working
PM=PR2=frac{sqrt{162+102}{2}=frac{sqrt{356}{2}=9.434 cm
tanθ=TMPM=159.434=1.590
Result
θ=tan-1(1.590)=57.9^circ
7.3.3.The diagram shows a pyramid. The base, PQRS, is a horizontal square of side 10 cm. The vertex, V, is vertically above the midpoint, M, and m VM = 12 cm. Calculate the size of angle VPM.
Working
PM=frac{diagonal of square{2}=10sqrt22=5sqrt2 cm
tanangle VPM=VMPM=125sqrt2=1.697
Result
angle VPM=tan-1(1.697)=59.3^circ
7.3.4.ABCDE is a square based pyramid, in which m AE = m BE = m CE = m DE = 12 cm and m AB = 15 cm. Calculate the size of angle DEB. Give your answer in degree (whole numbers).
7.3.5.The diagram shows a triangular prism with a horizontal rectangular base ABCD. m AB = 10 cm, m BC = 7 cm. M is the midpoint of AD. The vertex T is vertically above M and m MT = 6 cm. Calculate the size of the angle between TB and the base.
7.3.6.In an isometric game, the camera is placed at a 45° angle from the ground. If the player is 10 units in front and 10 units above, what is the direct line of sight distance?
Given
Check
tan 45^circ=1010=1 (checkmark)
Formula
Pythagoras
d2=102+102=200
Result
d=sqrt{200}=10sqrt2 units
7.3.7.A surveyor spots the top of a tower at an elevation angle of 28°. He is standing 60 metres from the base. Find the height of the tower.
Formula
tan 28^circ=h60
Working
h=60tan 28^circ=60(0.5317)
Result
h=31.90 m
7.3.8.A boat crosses a river 80 m wide, at a 60° angle to the current. What distance does the boat actually travel?
Formula
sin 60^circ=80d
Working
d=80sin 60^circ=160sqrt3
Result
d=92.38 m
7.3.9.A listener hears a sound from two speakers. One is 6 m directly ahead and the other is at 30° to the side, 6 m away. Find the distance between the speakers.
7.3.10.A 10 m ladder leans against a wall making an angle of 75° with the ground. How high does it reach up the wall?
Formula
sin 75^circ=h10
Working
h=10sin 75^circ=10(0.9659)
Result
h=9.66 m
7.3.11.A lighthouse is located on a cliff 80 m above sea level. A ship is spotted at an angle of depression of 12°. How far is the ship from the base of the cliff?
Formula
tan 12^circ=80x
Working
x=80tan 12^circ=800.2126
Result
x=376.35 m