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Geometry and Trigonometry

184 soru

Tüm türler184Çoktan seçmeli85Sayısal93Kısa cevap6
Soru 181Soru →

In triangle PQRPQR, the side lengths are p=8 cmp = 8\text{ cm} and q=15 cmq = 15\text{ cm}, and the included angle ∠R=60∘\angle R = 60^\circ. What is the length of side rr in centimeters?

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Cevap: 13

Cevap

The length of side rr is 13 cm13\text{ cm}.
Using the Cosine Rule formula r2=p2+q2−2pqcos⁡Rr^2 = p^2 + q^2 - 2pq \cos R with p=8p=8, q=15q=15, and R=60∘R=60^\circ, we calculate r2=64+225−240(0.5)=169r^2 = 64 + 225 - 240(0.5) = 169. Taking the square root gives r=13 cmr = 13\text{ cm}.

Adım Adım Çözüm

1
Identify the given values and appropriate trigonometric rule
Givens: p=8 cmp = 8\text{ cm}, q=15 cmq = 15\text{ cm}, ∠R=60∘\angle R = 60^\circ. Since two sides and the included angle (SAS) are given, use the Cosine Rule: r2=p2+q2−2pqcos⁡Rr^2 = p^2 + q^2 - 2pq \cos R.
The Cosine Rule is required to find the third side when two sides and their included angle are known.
2
Substitute the values into the Cosine Rule formula
r2=82+152−2(8)(15)cos⁡60∘r^2 = 8^2 + 15^2 - 2(8)(15) \cos 60^\circ
Replacing variables with their numerical equivalents sets up the algebraic calculation.
3
Calculate the terms and evaluate r2r^2
r2=64+225−240×0.5=289−120=169r^2 = 64 + 225 - 240 \times 0.5 = 289 - 120 = 169
Since cos⁡60∘=0.5\cos 60^\circ = 0.5, simplify the arithmetic operations.
4
Solve for side length rr
r=169=13 cmr = \sqrt{169} = 13\text{ cm}
Take the square root of both sides to obtain the length of side rr.

Anahtar Kavram

Applying the Cosine Rule to find an unknown side given two sides and the included angle (SAS)
Soru 182Soru →

In △ABC\triangle ABC, side b=10 cmb = 10\text{ cm}, side c=6 cmc = 6\text{ cm}, and ∠A=120∘\angle A = 120^\circ. What is the length of side aa in centimeters?

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Cevap: 14

Cevap

The length of side aa is 14 cm14\text{ cm}.
Applying the Cosine Rule a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A with b=10 cmb = 10\text{ cm}, c=6 cmc = 6\text{ cm}, and ∠A=120∘\angle A = 120^\circ gives a2=102+62−2(10)(6)(−0.5)=100+36+60=196a^2 = 10^2 + 6^2 - 2(10)(6)(-0.5) = 100 + 36 + 60 = 196. Taking the square root yields a=14 cma = 14\text{ cm}.

Adım Adım Çözüm

1
Identify the given parameters and select the relevant trigonometric rule
Given sides b=10 cmb = 10\text{ cm}, c=6 cmc = 6\text{ cm}, and included angle ∠A=120∘\angle A = 120^\circ. Use the Cosine Rule: a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A.
The Cosine Rule is required because two sides and the included angle (SAS) are given to find the third side.
2
Substitute the known values into the Cosine Rule equation
a2=102+62−2(10)(6)cos⁡(120∘)a^2 = 10^2 + 6^2 - 2(10)(6) \cos(120^\circ)
This sets up a single equation with the unknown side length aa.
3
Evaluate the cosine term and simplify the arithmetic expression
a2=100+36−120(−0.5)=136+60=196a^2 = 100 + 36 - 120(-0.5) = 136 + 60 = 196
Since 120∘120^\circ is an obtuse angle in the second quadrant, cos⁡(120∘)=−0.5\cos(120^\circ) = -0.5, changing the minus sign in the formula to a plus sign.
4
Calculate the principal square root to find aa
a=196=14 cma = \sqrt{196} = 14\text{ cm}
Side length must be a positive scalar quantity.

Anahtar Kavram

Cosine Rule for Side Length in Oblique Triangles
Soru 183Soru →

A trigonometric function is defined by y=Acos⁡(Bx)+Cy = A \cos(B x) + C, where A>0A > 0 and B>0B > 0. If the function has a maximum value of 77, a minimum value of −3-3, and a period of 120∘120^\circ, what is the value of A+B+CA + B + C?

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Cevap: 10

Cevap

The value of A+B+CA + B + C is 10.
Solving the system A+C=7A + C = 7 and −A+C=−3-A + C = -3 gives C=2C = 2 and A=5A = 5. Using the period relationship 120∘=360∘/B120^\circ = 360^\circ / B yields B=3B = 3. Adding these parameters together gives 5+3+2=105 + 3 + 2 = 10.

Adım Adım Çözüm

1
Calculate vertical shift C and amplitude A
C = 2, A = 5
For y = A cos(Bx) + C, Max = A + C and Min = -A + C. Thus C = (Max + Min)/2 = 2 and A = (Max - Min)/2 = 5.
2
Calculate period coefficient B
B = 3
The period T in degrees is given by 360° / B. With T = 120°, B = 360° / 120° = 3.
3
Compute the sum A + B + C
10
Sum the derived parameters: 5 + 3 + 2 = 10.

Anahtar Kavram

Deriving amplitude, period coefficient, and vertical shift from trigonometric graph properties
Soru 184Soru →

What is the smallest positive angle θ\theta, in degrees, that satisfies the trigonometric equation 3tan⁡(3θ)=3\sqrt{3}\tan(3\theta) = 3?

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Cevap: 20

Cevap

The smallest positive value of θ\theta is 20∘20^\circ.
Isolating tan⁡(3θ)\tan(3\theta) gives tan⁡(3θ)=33=3\tan(3\theta) = \frac{3}{\sqrt{3}} = \sqrt{3}. The principal acute angle whose tangent is 3\sqrt{3} is 60∘60^\circ. Equating 3θ=60∘3\theta = 60^\circ and solving for θ\theta yields θ=20∘\theta = 20^\circ.

Adım Adım Çözüm

1
Isolate the trigonometric function tan⁡(3θ)\tan(3\theta)
tan⁡(3θ)=3\tan(3\theta) = \sqrt{3}
Dividing both sides by \sqrt{3} isolates the tangent term and rationalizes the fraction to \sqrt{3}.
2
Determine the principal angle for 3θ3\theta
3θ=60∘3\theta = 60^\circ
The smallest positive angle whose tangent equals \sqrt{3} is 60^\circ.
3
Solve for θ\theta
θ=20∘\theta = 20^\circ
Dividing the principal angle 60^\circ by the coefficient 3 gives the smallest positive angle \theta.

Anahtar Kavram

Solving trigonometric equations with multiple angle arguments
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