hello! Can someone please solve this
using trig? I received some hints: SQR=SQB-RQB , Use the tan() angle subtraction formula on both angles, you’ll get equivalent equations both of them should involve h
+26387 Two pedestrians standing 10 feet away and 40 feet away from the base of a billboard structure
have equivalent views of the billboard,
meaning that the angles marked in the diagram are congruent.
If the billboard is 9 feet tall, what is its height h off the ground?
hello! Can someone please solve this
\(\text{Let $\angle SPR = \angle SQR = \alpha $ } \\ \text{Let $\angle RPQ = \beta $ } \\ \text{Let $\angle RQB = \gamma $ } \)
\(\begin{array}{|lrcll|} \hline 1. & \tan(\beta) &=& \dfrac{h}{40} \\ & \beta &=& \arctan\left(\dfrac{h}{40}\right) \\ \hline 2. &\tan(\gamma) &=& \dfrac{h}{10} \\ & \gamma &=& \arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline 3. & \tan(\alpha+\beta) &=& \dfrac{9+h}{40} \quad | \quad \beta = \arctan\left(\dfrac{h}{40}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{40}\right)\right) &=& \dfrac{9+h}{40} \\ & \alpha+\arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{40}\right) \\ & \alpha &=& \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) \\ \hline 4. & \tan(\alpha+\gamma)&=& \dfrac{9+h}{10} \quad | \quad \gamma = \arctan\left(\dfrac{h}{10}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{10}\right)\right)&=& \dfrac{9+h}{10} \\ & \alpha+\arctan\left(\dfrac{h}{10}\right)&=& \arctan\left(\dfrac{9+h}{10}\right) \\ & \alpha&=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline \text{Formula: } \arctan(x)-\arctan(y) = \arctan\left( \dfrac{x-y}{1+xy} \right) \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \alpha = \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left( \dfrac{ \dfrac{(9+h)}{40}-\dfrac{h}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } \right) &=& \arctan\left( \dfrac{ \dfrac{(9+h)}{10}-\dfrac{h}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \right) \\ \dfrac{ \dfrac{9}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } &=& \dfrac{ \dfrac{9}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \\\\ \dfrac{9}{40} \left( 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} \right) &=& \dfrac{9}{10} \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\\\ 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} &=& 4 \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\ \dfrac{100+(9+h)h}{100} &=& 4\left( \dfrac{1600+(9+h)h}{1600} \right) \\ \dfrac{100+(9+h)h}{100} &=& \dfrac{1600+(9+h)h}{400} \\ 4\left( 100+(9+h)h \right) &=& 1600+(9+h)h \\ 400+4(9+h)h &=& 1600+(9+h)h \\ 400 +36h+4h^2 &=& 1600+9h+h^2 \\ 3h^2 +27h-1200 &=& 0 \quad | \quad : 3 \\ h^2 +9h -400 &=& 0 \\ h &=& \dfrac{-9\pm \sqrt{81-4\cdot(-400)} }{2} \\ h &=& \dfrac{-9\pm \sqrt{1681} }{2} \\ h &=& \dfrac{-9\pm 41 }{2} \\ h &=& \dfrac{-9{\color{red}\mathbf{+}} 41 }{2} \quad | \quad h>0!\\ \mathbf{h} &\mathbf{=}& \mathbf{16} \\ \hline \end{array}\)
+26387 Two pedestrians standing 10 feet away and 40 feet away from the base of a billboard structure
have equivalent views of the billboard,
meaning that the angles marked in the diagram are congruent.
If the billboard is 9 feet tall, what is its height h off the ground?
hello! Can someone please solve this
\(\text{Let $\angle SPR = \angle SQR = \alpha $ } \\ \text{Let $\angle RPQ = \beta $ } \\ \text{Let $\angle RQB = \gamma $ } \)
\(\begin{array}{|lrcll|} \hline 1. & \tan(\beta) &=& \dfrac{h}{40} \\ & \beta &=& \arctan\left(\dfrac{h}{40}\right) \\ \hline 2. &\tan(\gamma) &=& \dfrac{h}{10} \\ & \gamma &=& \arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline 3. & \tan(\alpha+\beta) &=& \dfrac{9+h}{40} \quad | \quad \beta = \arctan\left(\dfrac{h}{40}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{40}\right)\right) &=& \dfrac{9+h}{40} \\ & \alpha+\arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{40}\right) \\ & \alpha &=& \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) \\ \hline 4. & \tan(\alpha+\gamma)&=& \dfrac{9+h}{10} \quad | \quad \gamma = \arctan\left(\dfrac{h}{10}\right) \\ & \tan\left(\alpha+\arctan\left(\dfrac{h}{10}\right)\right)&=& \dfrac{9+h}{10} \\ & \alpha+\arctan\left(\dfrac{h}{10}\right)&=& \arctan\left(\dfrac{9+h}{10}\right) \\ & \alpha&=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline \text{Formula: } \arctan(x)-\arctan(y) = \arctan\left( \dfrac{x-y}{1+xy} \right) \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \alpha = \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left(\dfrac{9+h}{40}\right)- \arctan\left(\dfrac{h}{40}\right) &=& \arctan\left(\dfrac{9+h}{10}\right)-\arctan\left(\dfrac{h}{10}\right) \\ \arctan\left( \dfrac{ \dfrac{(9+h)}{40}-\dfrac{h}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } \right) &=& \arctan\left( \dfrac{ \dfrac{(9+h)}{10}-\dfrac{h}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \right) \\ \dfrac{ \dfrac{9}{40} } { 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} } &=& \dfrac{ \dfrac{9}{10} } { 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} } \\\\ \dfrac{9}{40} \left( 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} \right) &=& \dfrac{9}{10} \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\\\ 1+ \dfrac{(9+h)}{10}\dfrac{h}{10} &=& 4 \left( 1+ \dfrac{(9+h)}{40}\dfrac{h}{40} \right) \\ \dfrac{100+(9+h)h}{100} &=& 4\left( \dfrac{1600+(9+h)h}{1600} \right) \\ \dfrac{100+(9+h)h}{100} &=& \dfrac{1600+(9+h)h}{400} \\ 4\left( 100+(9+h)h \right) &=& 1600+(9+h)h \\ 400+4(9+h)h &=& 1600+(9+h)h \\ 400 +36h+4h^2 &=& 1600+9h+h^2 \\ 3h^2 +27h-1200 &=& 0 \quad | \quad : 3 \\ h^2 +9h -400 &=& 0 \\ h &=& \dfrac{-9\pm \sqrt{81-4\cdot(-400)} }{2} \\ h &=& \dfrac{-9\pm \sqrt{1681} }{2} \\ h &=& \dfrac{-9\pm 41 }{2} \\ h &=& \dfrac{-9{\color{red}\mathbf{+}} 41 }{2} \quad | \quad h>0!\\ \mathbf{h} &\mathbf{=}& \mathbf{16} \\ \hline \end{array}\)
