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anemone
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Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
[sp]anemone said:Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
Well done Opalg (Yes) and thanks for participating!Opalg said:[sp]
If $QA=3$, $AB=5$ and $BQ=4$ then the other two sides of the triangle $PQR$ will both be $6\sqrt2$ units.Cosine rule in triangle $PQA$: $AP^2 = 9 + 72 - 2\cdot 3\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 45$.
Cosine rule in triangle $PRB$: $BP^2 = 16 + 72 - 2\cdot 4\cdot 6\sqrt2\cdot\dfrac1{\sqrt2} = 40$.
Cosine rule in triangle $APB$: $\cos\alpha = \dfrac{40 + 45 - 25}{2\cdot\sqrt{40}\cdot\sqrt{45}} = \dfrac1{\sqrt2}$.
Therefore $\alpha = 45^\circ$.[/sp]
anemone said:Let $PQR$ be a triangle with $\angle P=90^{\circ}$ and $PQ=PR$. Let $A$ and $B$ be points on the segment $QR$ such that $QA:AB:BR=3:5:4$. Find $\angle APB$.
Albert said:let point Q be the origin so we may construct the following :
Q(0,0),A(3,0),B(8,0),R(12,0) and P(6,6)
slop of PA=2
slop of PB=-3
and we have :tan ($\angle $ APB)=1
that is $\angle $APB=$45^o$
The formula for finding $\angle APB$ in a triangle with known side ratios is $ \angle APB = \cos^{-1} \left(\frac{a^2 + c^2 - b^2}{2ac}\right)$, where $a$, $b$, and $c$ are the side lengths of the triangle.
In order to determine which side ratios correspond to $\angle APB$, you can use the Law of Cosines. This law states that in a triangle with sides $a$, $b$, and $c$, the following equation holds: $c^2 = a^2 + b^2 - 2ab\cos\theta$, where $\theta$ is the angle opposite to side $c$. Therefore, by comparing the known side ratios to the equation, you can determine which side ratios correspond to $\angle APB$.
Yes, you can still find $\angle APB$ if only two of the side ratios are known. In this case, you can use the Law of Sines, which states that in a triangle with sides $a$, $b$, and $c$, the following proportion holds: $\frac{\sin\alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin\gamma}{c}$, where $\alpha$, $\beta$, and $\gamma$ are the angles opposite to sides $a$, $b$, and $c$, respectively. Using this proportion and the two known side ratios, you can solve for the third side and then use the Law of Cosines to find $\angle APB$.
Knowing $\angle APB$ in a triangle is significant because it allows you to determine the angles of the triangle, which can provide valuable information about the shape, size, and properties of the triangle. In addition, knowing the angles can help you to solve other problems involving the triangle, such as finding the area or perimeter.
Yes, there are some limitations to using side ratios to find $\angle APB$ in a triangle. One limitation is that the side ratios must be in proportion to the sides of the triangle, otherwise the formula will not work. In addition, the formula assumes that the triangle is a right triangle, so it may not be accurate in other types of triangles. Lastly, the formula does not take into account any external factors that may affect the triangle, such as measurement errors or deformities.