From the diagram below , x = t ( r - h ) / h
Further explanation
Firstly , let us learn about trigonometry in mathematics.
Suppose the ΔABC is a right triangle and ∠A is 90°.
sin ∠A = opposite / hypotenuse
cos ∠A = adjacent / hypotenuse
tan ∠A = opposite / adjacent
There are several trigonometric identities that need to be recalled, i.e.
\(cosec ~ A = \frac{1}{sin ~ A}\)
\(sec ~ A = \frac{1}{cos ~ A}\)
\(cot ~ A = \frac{1}{tan ~ A}\)
\(tan ~ A = \frac{sin ~ A}{cos ~ A}\)
Let us now tackle the problem!
Look at ΔADE in the attachment.
We will use the following formula to find relationship between variable t and h:
tan ∠A = opposite / adjacent
\(\tan \angle A = \frac{DE}{AD}\)
\(\large {\boxed{ \tan \angle A = \frac{h}{t} } }\) → Equation 1
Look at ΔABC in the attachment.
We will use the following formula to find relationship between variable r , t and x:
tan ∠A = opposite / adjacent
\(\tan \angle A = \frac{BC}{AB}\)
\(\large {\boxed{ \tan \angle A = \frac{r}{x + t} } }\) → Equation 2
Next we can substitute equation 1 to equation 2 :
\(\tan \angle A = \frac{r}{x+t}\)
\(\frac{h}{t} = \frac{r}{x+t}\)
\((x + t)h = r ~ t\)
\((x + t) = \frac{(r ~ t)}{h}\)
\(x = \frac{(r ~ t)}{h} - t\)
\(x = \frac{(r ~ t)}{h} - \frac{(h ~ t)}{h}\)
\(\large {\boxed {x = \frac{t(r - h)}{h}} }\)
Learn more
- Calculate Angle in Triangle : https://brainly.com/question/12438587
- Periodic Functions and Trigonometry : https://brainly.com/question/9718382
- Trigonometry Formula : https://brainly.com/question/12668178
Answer details
Grade: College
Subject: Mathematics
Chapter: Trigonometry
Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse , Triangle , Fraction , Lowest , Function , Angle
