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Channel Posts
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2021 WASSCE MATHS QUESTION... SOLUTIONS FROM MY ARENA!!!
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| 2 |  I shall drop the remaining Confirmed topics only in' my Wattsap WASSCE vip platform β
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| 9 | +3 No text... | 39 |
| 10 | Practise these questions given here to get a deep knowledge of Trigonometry. Use the formulas and table given in this article wherever necessary.πππ
Q.1: In β³ABC, right-angled at B, AB=22 cm and BC=17 cm. Find:
(a) sin A Cos B
(b) tan A tan B
Q.2: If 12cot ΞΈ= 15, then find sec ΞΈ.
Q.3: In Ξ PQR, right-angled at Q, PR + QR = 30 cm and PQ = 10 cm. Determine the values of sin P, cos P and tan P.
Q.4: If sec 4ΞΈ = cosec (ΞΈ- 300), where 4ΞΈ is an acute angle, find the value of A.β
β
β
| 76 |
| 11 | Example 4:
Find the value of sin 75β°.
Solution:
We can write,
sin 75β° = sin (45β° + 30β°) = sin 45β° cos 30β°+ cos 45β° sin 30β°
{Note please, since, sin (A + B) = sin A cos B + cos A sin B}
= 1/β2 Γ β3/2 + 1/β2 Γ 1/2 = (β3 + 1)/2β2
β΄ sin 75β°= (β3 + 1)/2β2
Are you clear???...show reaction βπ½ | 64 |
| 12 | Example 3: Find out the exact value of sin15ΒΊ using the trig value for standard angles from a trigonometry table.
Solution:
Using the trigonometric values table, we know that sin 45ΒΊ = 1/β2, cos 30ΒΊ = (β3/2), cos 45ΒΊ = 1/β2, and sin 30ΒΊ = 1/2
sin 15ΒΊ = sin(45ΒΊ - 30ΒΊ) = sin45ΒΊ cos 30ΒΊ - cos 45ΒΊsin 30ΒΊ = (β2/2) β’ (β3/2) - (β2/2) β’ (1/2) = (β6 - β2)/4 = (β3 - 1)/2β2
Answer: The value of sin15ΒΊ = (β3 - 1)/2β2
MR OYIBO CEO βπ½βπ½ | 45 |
| 13 | Orπππ
We can further resolve the above-resulted expression by putting the value of β3, which is equal to 1.732.
β΄ Tan (15Β°) = (1.732 β 1)/(1.732 + 1) = 0.2679
Or tan (15Β°) β 0.27
Tan 15Β° With Respect To sin and cos function
Similarly, we can also find the value of tangent 15 degrees, by knowing the value of sin 15 and cos 15 degrees.
Tan (15Β°) = sin 15/cos 15
Tan 15Β° = sin 15/cos 15
Sin 15Β° = sin (45 β 30)Β° and cos 15 = cos (45 β 30)Β°
β΄ tan (15Β°) =Β sin (45 β 30)Β° /cos (45 β 30)Β°
From the trigonometry formulas, we know,
sin(A β B) = sin A cos B β cos A sin B
and cos (A β B) = cos A cos B + sin A sin B
Therefore,
tan (15Β°)= (sin 45Β° cos 30Β° β cos 45Β° sin 30Β°)/ (cos 45Β° cos 30Β° + sin 45Β° sin 30Β°)
Putting the values of sin 30Β°, sin 45Β°, cos 30Β° and cos 45Β°, we get,
tan 15Β° = [(1/β2).(β3/2) β (1/β2).(Β½)] / [(1/β2).(β3/2) + (1/β2).(Β½)]
Solving the above equation we have,
tan 15Β° = β3 β 1/ β3 + 1 | 42 |
| 14 | From the above table, we have the values of tan, sin and cos ratios for 0Β°, 30Β°, 45Β°, 60Β° and 90Β°. Now, by using these values, we have to find the value of tan (15Β°). Letβs get started,πππ
Tan 15Β° = Tan(45 β 30)Β°
By the trigonometry formula, we know,β
ππ
Tan (A β B) = (Tan A β Tan B) /(1 + Tan A Tan B)
Therefore, we can write,
tan(45 β 30)Β° = tan 45Β° β tan 30Β°/1+tan 45Β° tan 30Β°
Now putting the values of tan 45Β° and tan 30Β° from the table we get;πππ
tan(45 β 30)Β° = (1 β 1/β3)/ (1 + 1.1/β3)
tan (15Β°) = β3 β 1/ β3 + 1
Hence, the value of tan (15Β°) is β3 β 1/β3 + 1.β
β
| 33 |
| 15 | β
| 33 |
| 16 | Example 2. find the value of tan 15β°, with the help of trigonometry table.β
β
Before we try to find the tan 15 degrees value, letβs have a look at trigonometry table for sin, cos and tan.πππ | 37 |
| 17 | Ok two moreππ | 43 |
| 18 | Solved Examples
Example 1:
Find the value of tan 45β° + 2 cos 60β° β sec 60β°.
Solution:
From the trigonometry table, tan 45β° = 1, cos 60β° = Β½ and sec 60β° = 2
Therefore, tan 45β° + 2 cos 60β° β sec 60β° = 1 + 2 Γ Β½ β 2 = 1 + 1 β 2 = 0
SIR OYIBO(THEMAINMAN)
βπ½βπ½βπ½ | 46 |
| 19 | β
β
π | 38 |
| 20 | Step 7: Determine the value of sec
The value of sec can be determined by all reciprocal values of cos. The value of sec on 0Β° is the opposite of cos on 0Β°. So the value will be:
In the same way, the table for sec is given below.ππ | 33 |
