Proving Triangle Angles Relation: $\frac{1}{\sin^2 \lambda}$

In summary, the equation for proving triangle angle relations using $\frac{1}{\sin^2 \lambda}$ involves calculating the sum of the squares of the reciprocals of the trigonometric functions of the angles in a triangle. This equation is commonly used in surveying, navigation, engineering, and construction. However, it can only be used for right triangles and assumes accurate measurements.
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anemone
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Inside a triangle $ABC$, there is a point P satisfies $\angle PAB=\angle PBC=\angle PCA=\lambda$. If the angles of the triangle are denoted by $\alpha$, $\beta$ and $\gamma$, prove that

$\dfrac{1}{\sin^2 \lambda}=\dfrac{1}{\sin^2 \alpha}+\dfrac{1}{\sin^2 \beta}+\dfrac{1}{\sin^2 \gamma}$
 
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anemone said:
Inside a triangle $ABC$, there is a point P satisfies $\angle PAB=\angle PBC=\angle PCA=\lambda$. If the angles of the triangle are denoted by $\alpha$, $\beta$ and $\gamma$, prove that

$\dfrac{1}{\sin^2 \lambda}=\dfrac{1}{\sin^2 \alpha}+\dfrac{1}{\sin^2 \beta}+\dfrac{1}{\sin^2 \gamma}$
using the sine law:
View attachment 1935
 

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Last edited:
  • #3
Thanks for participating, Albert!

Solution provided by other:
View attachment 1936
First we let AP meet BC at X.

Since $\angle XBP=\angle BAX=\lambda$ and $\angle BXP=\angle AXB=\text{common angle}$, we can say that triangle XPB and XBA are similar. Then we have $\dfrac{XB}{XP}=\dfrac{XA}{XB}$

Applying the sine law to the triangle XBA and considering the fact that $\dfrac{XB}{XP}=\dfrac{XA}{XB}$ give us

$\dfrac{\sin \lambda}{\sin \beta}=\dfrac{XB}{XA}$

$\dfrac{\sin^2 \lambda}{\sin^2 \beta}=\dfrac{XB^2}{XA^2}=\dfrac{XAXP}{XA^2}= \dfrac{XP}{XA}$

Now, the ratio of the area of the triangles XPB to XBA and triangles XCP to XCA are

$\dfrac{\text{area of the triangle XPB}}{\text{area of the triangle XBA}}=\dfrac{XP}{XA}$ and

$\dfrac{\text{area of the triangle XCP}}{\text{area of the triangle XCA}}=\dfrac{XP}{XA}$

This tells us $\dfrac{\text{area of the triangle XPB}}{\text{area of the triangle XBA}}=\dfrac{\text{area of the triangle XCP}}{\text{area of the triangle XCA}}=\dfrac{\text{area of the triangle BPC}}{\text{area of the triangle ABC}}$

$\therefore \dfrac{\sin^2 \lambda}{\sin^2 \beta}=\dfrac{\text{area of the triangle BPC}}{\text{area of the triangle ABC}}$

By the similar arguments, we have

$\therefore \dfrac{\sin^2 \lambda}{\sin^2 \alpha}=\dfrac{\text{area of the triangle APB}}{\text{area of the triangle ABC}}$

$\therefore \dfrac{\sin^2 \lambda}{\sin^2 \gamma}=\dfrac{\text{area of the triangle CPA}}{\text{area of the triangle ABC}}$

Hence,

$\small\dfrac{\sin^2 \lambda}{\sin^2 \alpha}+\dfrac{\sin^2 \lambda}{\sin^2 \beta}+\dfrac{\sin^2 \lambda}{\sin^2 \gamma}=\dfrac{\text{area of the triangle APB}}{\text{area of the triangle ABC}}+\dfrac{\text{area of the triangle BPC}}{\text{area of the triangle ABC}}+\dfrac{\text{area of the triangle CPA}}{\text{area of the triangle ABC}}=1$

$\dfrac{1}{\sin^2 \lambda}=\dfrac{1}{\sin^2 \alpha}+\dfrac{1}{\sin^2 \beta}+\dfrac{1}{\sin^2 \gamma}$ (Q.E.D.)
 

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FAQ: Proving Triangle Angles Relation: $\frac{1}{\sin^2 \lambda}$

What is the equation for proving triangle angle relations using $\frac{1}{\sin^2 \lambda}$?

The equation is $\frac{1}{\sin^2 \lambda} = \frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2}$, where $a,b,c,d,e,f$ represent the sides of the triangle.

What does $\lambda$ represent in the equation?

$\lambda$ represents the angle opposite side $a$ in the triangle.

How is this equation used to prove triangle angle relations?

This equation is used to calculate the sum of the squares of the reciprocals of the trigonometric functions of the angles in a triangle. If the sum is equal to 1, then the triangle is a right triangle.

What are some real-life applications of using $\frac{1}{\sin^2 \lambda}$ to prove triangle angle relations?

This equation is commonly used in surveying and navigation to calculate the angles and distances between points on a map or in the field. It is also used in engineering and construction to ensure the accuracy of angles in building structures.

Are there any limitations to using $\frac{1}{\sin^2 \lambda}$ to prove triangle angle relations?

Yes, this equation can only be used for right triangles. It also assumes that the sides and angles of the triangle are measured accurately, and any measurement errors can affect the accuracy of the results.

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