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Trigonometry: Area of Triangles

Based on the books answers, **(i) **& **(ii)** are correct but I'm confused about **(ii)**, seeing as how the answer is provided in the question!! Am I meant to prove the statement through the information provided...?

The book's answer to **(iii)** is provided below, but the confusion for me here is the application of tan, in defining the area of the triangle. The 1st attempt in **(iii)** is what I expected the answer to be but, based on the books final answer, I made a second attempt to try and account for tan. Also, again based on the books answer, I'm assuming |ad| is meant to = r but I'm unsure as to how I prove this. Can anyone help me out here?

Many thanks.

Q. A triangle is inscribed in a sector of a circle, center c, radius r, $\displaystyle \theta<90^o$. A right-angled triangle cad circumscribes the sector, as shown. If the area of a sector is $\displaystyle \frac{1}{2}r^2\theta$, find the area of **(i)** $\displaystyle \triangle cab$, **(ii)** the sector cab, **(iii)** $\displaystyle \triangle$ cad. Hence show that $\displaystyle sin\theta<\theta<tan\theta$.

**Attempt: (i) **Area of $\displaystyle \triangle abc$: $\displaystyle \frac{1}{2}(a)(b)(sinC)$

|ac| = r = |bc|, where both lines are connected from the center to the circumference of the circle whose sector is cab

Thus, $\displaystyle \frac{1}{2}(a)(b)(sinC)$ = $\displaystyle \frac{1}{2}r^2sin\theta$

**(ii) **$\displaystyle \frac{1}{2}r^2\theta$

**(iii) **Area of $\displaystyle \triangle cad$: $\displaystyle \frac{1}{2}(d)(a)(sinA)$ = $\displaystyle \frac{1}{2}(r)(|cd|)(sin90^o)$ = $\displaystyle \frac{1}{2}(r)(a)(1)$ = $\displaystyle \frac{1}{2}ra$

**or...?**

$\displaystyle \frac{1}{2}(c)(d)(tan\theta)$ = $\displaystyle \frac{1}{2}(|ab|)(r)(tan\theta)$

Ans.: (From text book): (iii) $\displaystyle \frac{1}{2}r^2tan\theta$

Re: Trigonometry: Area of Triangles

note that $\displaystyle |ad| = r\tan{\theta}$

$\displaystyle A = \frac{1}{2}bh = \frac{1}{2} r \cdot r\tan{\theta}$

Re: Trigonometry: Area of Triangles

Thanks.

Ok, so if I'm to demonstrate $\displaystyle sin\theta<\theta<tan\theta$, and knowing that $\displaystyle \theta<90^o$, I can simply pick any degree from $\displaystyle 0^0-89^0$ and display the results e.g. $\displaystyle \theta=60^0$ in which case $\displaystyle sin\60^o<\60^o<tan\60^0$ = $\displaystyle 0.86<60<1.73$. Although, this leaves me with the problem of the 60 in the middle.

Is there a better way to prove the statement?

Re: Trigonometry: Area of Triangles

Quote:

Originally Posted by

**GrigOrig99** Thanks.

Ok, so if I'm to demonstrate $\displaystyle sin\theta<\theta<tan\theta$, and knowing that $\displaystyle \theta<90^o$, I can simply pick any degree from $\displaystyle 0^0-89^0$ and display the results e.g. $\displaystyle \theta=60^0$ in which case $\displaystyle sin\60^o<\60^o<tan\60^0$ = $\displaystyle 0.86<60<1.73$. Although, this leaves me with the problem of the 60 in the middle.

Is there a better way to prove the statement?

$\displaystyle \text{Area}(\Delta CAB)\le \text{Sector-area}(CAB)\le\text{Area}(\Delta CAD) $

Re: Trigonometry: Area of Triangles