Math Problem Statement

Right-angled triangles with missing trigonometric values. Solve for sin, cos, tan, and their reciprocal functions for each triangle.

Solution

The image shows several right-angled triangles with missing trigonometric values. Let's solve each triangle step by step using trigonometric relationships.

1. Triangle ABC:

  • Given: AB=8AB = 8, BC=10BC = 10, B=90\angle B = 90^\circ
  • To find: sinA,cosA,tanA,cscA,secA,cotAsinA, cosA, tanA, cscA, secA, cotA

Using the Pythagorean theorem: AC=AB2+BC2=82+102=64+100=16412.81AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164} \approx 12.81

Now, we can find the trigonometric values:

  • sinA=BCAC=1012.810.780sinA = \frac{BC}{AC} = \frac{10}{12.81} \approx 0.780
  • cosA=ABAC=812.810.625cosA = \frac{AB}{AC} = \frac{8}{12.81} \approx 0.625
  • tanA=BCAB=108=1.25tanA = \frac{BC}{AB} = \frac{10}{8} = 1.25
  • cscA=1sinA=10.7801.282cscA = \frac{1}{sinA} = \frac{1}{0.780} \approx 1.282
  • secA=1cosA=10.625=1.6secA = \frac{1}{cosA} = \frac{1}{0.625} = 1.6
  • cotA=1tanA=11.25=0.8cotA = \frac{1}{tanA} = \frac{1}{1.25} = 0.8

2. Triangle PQR:

  • Given: PQ=15PQ = 15, PR=8PR = 8, R=90\angle R = 90^\circ
  • To find: sinQ,cosQ,tanQ,cscQ,secQ,cotQsinQ, cosQ, tanQ, cscQ, secQ, cotQ

Using the Pythagorean theorem: QR=PQ2PR2=15282=22564=16112.69QR = \sqrt{PQ^2 - PR^2} = \sqrt{15^2 - 8^2} = \sqrt{225 - 64} = \sqrt{161} \approx 12.69

Now, we can find the trigonometric values:

  • sinQ=PRPQ=815=0.533sinQ = \frac{PR}{PQ} = \frac{8}{15} = 0.533
  • cosQ=QRPQ=12.69150.846cosQ = \frac{QR}{PQ} = \frac{12.69}{15} \approx 0.846
  • tanQ=PRQR=812.690.631tanQ = \frac{PR}{QR} = \frac{8}{12.69} \approx 0.631
  • cscQ=1sinQ=10.5331.876cscQ = \frac{1}{sinQ} = \frac{1}{0.533} \approx 1.876
  • secQ=1cosQ=10.8461.182secQ = \frac{1}{cosQ} = \frac{1}{0.846} \approx 1.182
  • cotQ=1tanQ=10.6311.585cotQ = \frac{1}{tanQ} = \frac{1}{0.631} \approx 1.585

3. Triangle XYZ:

  • Given: XY=4XY = 4, YZ=1YZ = 1, Y=90\angle Y = 90^\circ
  • To find: sinX,cosX,tanX,cscX,secX,cotXsinX, cosX, tanX, cscX, secX, cotX

Using the Pythagorean theorem: XZ=XY2+YZ2=42+12=16+1=174.123XZ = \sqrt{XY^2 + YZ^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \approx 4.123

Now, we can find the trigonometric values:

  • sinX=YZXZ=14.1230.242sinX = \frac{YZ}{XZ} = \frac{1}{4.123} \approx 0.242
  • cosX=XYXZ=44.1230.970cosX = \frac{XY}{XZ} = \frac{4}{4.123} \approx 0.970
  • tanX=YZXY=14=0.25tanX = \frac{YZ}{XY} = \frac{1}{4} = 0.25
  • cscX=1sinX=10.2424.132cscX = \frac{1}{sinX} = \frac{1}{0.242} \approx 4.132
  • secX=1cosX=10.9701.031secX = \frac{1}{cosX} = \frac{1}{0.970} \approx 1.031
  • cotX=1tanX=10.25=4cotX = \frac{1}{tanX} = \frac{1}{0.25} = 4

4. Triangle RST:

  • Given: RS=23RS = \frac{\sqrt{2}}{3}, S=90\angle S = 90^\circ
  • To find: sinR,cosR,tanR,cscR,secR,cotRsinR, cosR, tanR, cscR, secR, cotR

Without the second side, it is hard to directly calculate trigonometric values for this triangle. If additional information is provided, I can proceed.


Would you like further clarifications or details on any of these solutions?

Related Questions:

  1. How is the Pythagorean theorem applied in right triangles?
  2. What are the reciprocal trigonometric functions, and how are they calculated?
  3. How do we determine angles using trigonometric functions if the sides are known?
  4. Why is the tangent of an angle the ratio of the opposite and adjacent sides?
  5. How do we approximate square roots in trigonometric problems?

Tip:

For quick calculations of trigonometric functions, remember that calculators typically provide sine, cosine, and tangent, while their reciprocals need manual computation.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right-Angled Triangles
Reciprocal Trigonometric Functions

Formulas

sin(θ) = opposite / hypotenuse
cos(θ) = adjacent / hypotenuse
tan(θ) = opposite / adjacent
csc(θ) = 1 / sin(θ)
sec(θ) = 1 / cos(θ)
cot(θ) = 1 / tan(θ)

Theorems

Pythagorean Theorem: a^2 + b^2 = c^2

Suitable Grade Level

Grades 9-12