Question to test your Inverse Trigonometric Functions knowledge

Question

$ABC$ is a right-angled triangle as shown below.

Right-angled triangle ABC with sides y and 15, used to find angle C in inverse trigonometric functions.

Find the measure of angle $C$ in terms of the side " $y$ ".

Write the answer using $arc\cos$ , $arc\cot$ , and $arc\csc$ .

Accepted Answer

﻿<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mrow><mi>cos</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt><mn>15</mn></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>cot</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt><mi>y</mi></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>csc</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mn>15</mn><mi>y</mi></mfrac><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">\cos^{-1}\left(\frac{\sqrt{225-y^2}}{15}\right)=\cot^{-1}\left(\frac{\sqrt{225-y^2}}{y}\right)=\csc^{-1}\left(\frac{15}{y}\right)</annotation></semantics></math>cos−1(15225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​​)=cot−1(y225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​​)=csc−1(y15​)﻿

Answer

﻿<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mrow><mi>cos</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mi>y</mi><mn>15</mn></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>cot</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt><mi>y</mi></mfrac><mo fence="true">)</mo></mrow><mo>−</mo><mi>c</mi><mi>s</mi><msup><mi>c</mi><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mn>15</mn><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt></mfrac><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">\cos^{-1}\left(\frac{y}{15}\right)=\cot^{-1}\left(\frac{\sqrt{225-y^2}}{y}\right)-csc^{-1}\left(\frac{15}{\sqrt{225-y^2}}\right)</annotation></semantics></math>cos−1(15y​)=cot−1(y225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​​)−csc−1(225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​15​)﻿

Answer

﻿<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mrow><mi>cos</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt><mn>15</mn></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>cot</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mi>y</mi><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>csc</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mn>15</mn><mi>y</mi></mfrac><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">\cos^{-1}\left(\frac{\sqrt{225-y^2}}{15}\right)=\cot^{-1}\left(\frac{y}{\sqrt{225-y^2}}\right)=\csc^{-1}\left(\frac{15}{y}\right)</annotation></semantics></math>cos−1(15225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​​)=cot−1(225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​y​)=csc−1(y15​)﻿

Answer

﻿<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mrow><mi>cos</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt><mn>15</mn></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>cot</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mi>y</mi><msqrt><mrow><mn>225</mn><mo>−</mo><msup><mi>y</mi><mn>2</mn></msup></mrow></msqrt></mfrac><mo fence="true">)</mo></mrow><mo>=</mo><msup><mrow><mi>csc</mi><mo>⁡</mo></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo fence="true">(</mo><mfrac><mi>y</mi><mn>15</mn></mfrac><mo fence="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">\cos^{-1}\left(\frac{\sqrt{225-y^2}}{15}\right)=\cot^{-1}\left(\frac{y}{\sqrt{225-y^2}}\right)=\csc^{-1}\left(\frac{y}{15}\right)</annotation></semantics></math>cos−1(15225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​​)=cot−1(225−y2<svg width="400em" height="1.5429em" viewBox="0 0 400000 1080" preserveAspectRatio="xMinYMin slice"><path d="M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z"></path></svg>​y​)=csc−1(15y​)﻿

Inverse Trigonometric Functions: Study with Video Lessons, Practice Problems & Examples

﻿ABCABCABC﻿ is a right-angled triangle as shown below.Find the measure of angle ﻿CCC﻿ in terms of the side "﻿yyy﻿".Write the answer using ﻿arccos⁡arc\cosarccos﻿, ﻿arccot⁡arc\cotarccot﻿, and ﻿arccsc⁡arc\cscarccsc﻿.

$ABC$ is a right-angled triangle as shown below.
Find the measure of angle $C$ in terms of the side " $y$ ".
Write the answer using $arc\cos$ , $arc\cot$ , and $arc\csc$ .