Saoirse29idn
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BALE 20 POINTS TO PAGNASAGUTAN MO NG TAMA SAGUTAN MO RIN TO SA ISA KO KAHIT ANO NALANG SAGOT MO PAG NASAGOT MO TO REPORT AGAD PAGKALOKOHAN LANG.

You focus on camera on a fountain. Your camera is at the vertex of angle formed by tangents to the fountain.You estate that this angle is 40°. What will the measure of the arc of the circular basin of the fountain that will he in photograph? ​

BALE 20 POINTS TO PAGNASAGUTAN MO NG TAMA SAGUTAN MO RIN TO SA ISA KO KAHIT ANO NALANG SAGOT MO PAG NASAGOT MO TO REPORT AGAD PAGKALOKOHAN LANGYou Focus On Came class=

Sagot :

EngrJohannidn

[tex] \large \bold{SOLUTION:} [/tex]

We know that the measure of an external angle or outside angle of the circle, formed by two tangent segments, is half the difference of the measures of its intercepted arcs. Let [tex]C[/tex] be a point on the larger arc intercepted by an external angle 40°.

Given that [tex]\textsf{m}\overset{\Large \frown}{AB} = x\degree [/tex],

[tex] \begin{array}{l} \begin{aligned} \qquad 40\degree &= \frac{1}{2}(\textsf{m}\overset{\huge \frown}{ACB} - \textsf{m}\overset{\Large \frown}{AB}) \\ 40\degree &= \frac{1}{2}\big[(360 - x)\degree - x\degree\big] \\ 40\degree &= \frac{1}{2}(360 - 2x)\degree \\ 40 &= 180 - x \\ x &= 180 - 40 \\ x &= 140 \end{aligned} \\ \\ \implies \textsf{m}\overset{\Large \frown}{AB} = x\degree = \boxed{140\degree}\quad\textit{Answer} \end{array} [/tex]

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