Let quadrilateral ABCD satisfy $\mathrm{\angle}BAC=\mathrm{\angle}CAD=2\phantom{\rule{thinmathspace}{0ex}}\mathrm{\angle}ACD={40}^{\circ}$ and $\mathrm{\angle}ACB={70}^{\circ}$. Find $\mathrm{\angle}ADB$.

Ximena Skinner
2022-07-15
Answered

Let quadrilateral ABCD satisfy $\mathrm{\angle}BAC=\mathrm{\angle}CAD=2\phantom{\rule{thinmathspace}{0ex}}\mathrm{\angle}ACD={40}^{\circ}$ and $\mathrm{\angle}ACB={70}^{\circ}$. Find $\mathrm{\angle}ADB$.

You can still ask an expert for help

Dalton Lester

Answered 2022-07-16
Author has **12** answers

Step 1

$\approx 77,{34}^{\circ}$, so

Step 2

$\frac{\mathrm{sin}80}{\mathrm{sin}40}\frac{\mathrm{sin}(70-x)}{\mathrm{sin}x}\frac{\mathrm{sin}20}{\mathrm{sin}90}=1$

After some manipulation we get $\mathrm{cot}x=\mathrm{tan}20+\frac{2}{\mathrm{cos}10}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=...$

$\approx 77,{34}^{\circ}$, so

Step 2

$\frac{\mathrm{sin}80}{\mathrm{sin}40}\frac{\mathrm{sin}(70-x)}{\mathrm{sin}x}\frac{\mathrm{sin}20}{\mathrm{sin}90}=1$

After some manipulation we get $\mathrm{cot}x=\mathrm{tan}20+\frac{2}{\mathrm{cos}10}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=...$

Joel French

Answered 2022-07-17
Author has **10** answers

Since $\mathrm{\angle}ACB=\mathrm{\angle}ABC={70}^{\circ}$, the triangle ABC is isosceles and $\phantom{\rule{thickmathspace}{0ex}}\overline{AB}=\overline{AC}$.

By applying the law of sines to the triangle ACD, we get that:

$\overline{AD}=\overline{AC}\cdot \frac{{\textstyle \mathrm{sin}\mathrm{\angle}ACD}}{{\textstyle \mathrm{sin}\mathrm{\angle}ADC}}=\overline{AC}\cdot \frac{{\textstyle \mathrm{sin}{20}^{\circ}}}{{\textstyle \mathrm{sin}{120}^{\circ}}}=\frac{{\textstyle 2\overline{AC}\mathrm{sin}{20}^{\circ}}}{{\textstyle \sqrt{3}}}\phantom{\rule{thickmathspace}{0ex}}.$

And, by applying the law of sines to the triangle ABD, we get that:

$\overline{AD}\mathrm{sin}\mathrm{\angle}ADB=\overline{AB}\mathrm{sin}\mathrm{\angle}ABD\phantom{\rule{thickmathspace}{0ex}}.\phantom{\rule{1em}{0ex}}{(\ast )}$

Let $\phantom{\rule{thickmathspace}{0ex}}\alpha =\mathrm{\angle}ADB\phantom{\rule{thickmathspace}{0ex}}.$.

Step 2

Since $\phantom{\rule{thickmathspace}{0ex}}\overline{AD}=\frac{{\textstyle 2\overline{AC}\mathrm{sin}{20}^{\circ}}}{{\textstyle \sqrt{3}}}\phantom{\rule{thickmathspace}{0ex}}$, $\phantom{\rule{thickmathspace}{0ex}}\overline{AB}=\overline{AC}\phantom{\rule{thickmathspace}{0ex}}$ and $\phantom{\rule{thickmathspace}{0ex}}\mathrm{\angle}ABD={100}^{\circ}-\alpha \phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}$, the equality (∗) turns into:

$\frac{{\textstyle 2\overline{AC}\mathrm{sin}{20}^{\circ}\mathrm{sin}\alpha}}{{\textstyle \sqrt{3}}}=\overline{AC}\mathrm{sin}({100}^{\circ}-\alpha )\phantom{\rule{thickmathspace}{0ex}},$

$2\mathrm{sin}{20}^{\circ}\mathrm{sin}\alpha =\sqrt{3}\mathrm{sin}({90}^{\circ}+{10}^{\circ}-\alpha )\phantom{\rule{thickmathspace}{0ex}},$

$4\mathrm{sin}{10}^{\circ}\mathrm{cos}{10}^{\circ}\mathrm{sin}\alpha =\sqrt{3}\mathrm{cos}({10}^{\circ}-\alpha )\phantom{\rule{thickmathspace}{0ex}},$

$4\mathrm{sin}{10}^{\circ}\mathrm{cos}{10}^{\circ}\mathrm{sin}\alpha =\sqrt{3}(\mathrm{cos}{10}^{\circ}\mathrm{cos}\alpha +\mathrm{sin}{10}^{\circ}\mathrm{sin}\alpha )\phantom{\rule{thickmathspace}{0ex}},$

$4\mathrm{sin}{10}^{\circ}\mathrm{sin}\alpha =\sqrt{3}(\mathrm{cos}\alpha +\mathrm{tan}{10}^{\circ}\mathrm{sin}\alpha )\phantom{\rule{thickmathspace}{0ex}},$

$(4\mathrm{sin}{10}^{\circ}-\sqrt{3}\mathrm{tan}{10}^{\circ})\mathrm{sin}\alpha =\sqrt{3}\mathrm{cos}\alpha \phantom{\rule{thickmathspace}{0ex}},$

$\mathrm{tan}\alpha =\frac{{\textstyle \sqrt{3}}}{{\textstyle 4\mathrm{sin}{10}^{\circ}-\sqrt{3}\mathrm{tan}{10}^{\circ}}}\phantom{\rule{thickmathspace}{0ex}}.$

Hence, $\mathrm{\angle}ADB=\alpha =\mathrm{arctan}\left(\frac{{\textstyle \sqrt{3}}}{{\textstyle 4\mathrm{sin}{10}^{\circ}-\sqrt{3}\mathrm{tan}{10}^{\circ}}}\right)\simeq \phantom{\rule{0ex}{0ex}}\simeq 77,{3361794}^{\circ}.$

asked 2022-08-25

What is the measure of $\measuredangle x$ in the figure below?

For reference: In the figure shown P and Q are points of tangency and $\stackrel{\u2322}{BC}+\stackrel{\u2322}{FE}={130}^{o}.$. Calculate "x".

For reference: In the figure shown P and Q are points of tangency and $\stackrel{\u2322}{BC}+\stackrel{\u2322}{FE}={130}^{o}.$. Calculate "x".

asked 2022-07-14

We have $AB=BC$, $AC=CD$, $\mathrm{\angle}ACD={90}^{\circ}$. If the radius of the circle is 'r' , find BC in terms of r .

asked 2022-08-13

Prove that the intersection angle between the Simson lines of two triangles inscribed in the same circle it's the same for any point.

Suppose the triangles ABC and DEF share the circuncircle C, and P and Q are any diferent points on C. Let l, m be, the Simson lines of P related to ABC and DEF, and p, q the Simson lines of Q related to ABC and DEF, then i must prove the angle between l and m equals the angle between p and q.

I just have one Theorem about the Simson line:

Theorem: Let P, Q be two points on the circuncircle, C, of the triangle ABC. Let l, m be their respective Simson's lines. Then the angle between l an m equals to the half of the central angle POQ, where O is the center of C.

Suppose the triangles ABC and DEF share the circuncircle C, and P and Q are any diferent points on C. Let l, m be, the Simson lines of P related to ABC and DEF, and p, q the Simson lines of Q related to ABC and DEF, then i must prove the angle between l and m equals the angle between p and q.

I just have one Theorem about the Simson line:

Theorem: Let P, Q be two points on the circuncircle, C, of the triangle ABC. Let l, m be their respective Simson's lines. Then the angle between l an m equals to the half of the central angle POQ, where O is the center of C.

asked 2022-07-25

For a triangle ABC, prove that angle C is a right angle if and only if the median of side c is c/2

asked 2022-07-22

In the following figure, $AD=AB$. Also $\mathrm{\angle}DAB=\mathrm{\angle}DCB=\mathrm{\angle}AEC={90}^{\circ}$ and $AE=5$. Find the area of quadrilateral ABCD.

$\sqrt{({x}^{2}-25)}$

$\sqrt{({x}^{2}-25)}$

asked 2022-09-12

the perimeter of the rhombus is 40. and one of the angles is 30 degrees. find the area of the rhombus

asked 2022-07-27

What's an example of a true conditional statement with a true converse?