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triangle a: 53, 71, 124; triangle b: 53, 71, 124 triangle c: 90, 37, __…

Question

triangle a: 53, 71, 124; triangle b: 53, 71, 124
triangle c: 90, 37, __; triangle d: 90, 53, __
triangle e: 63, 45, __; triangle f: 14, 71, __
triangle g: 121, __, ; triangle h: 70, , __

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of the interior angles of a triangle is 180°.

Step2: For Triangle C

Let the unknown angle be $x$. Then $x = 180-(90 + 37)=180 - 127 = 53$.

Step3: For Triangle D

Let the unknown angle be $y$. Then $y=180-(90 + 53)=180 - 143 = 37$.

Step4: For Triangle E

Let the unknown angle be $z$. Then $z = 180-(63 + 45)=180 - 108 = 72$.

Step5: For Triangle F

Let the unknown angle be $w$. Then $w=180-(14 + 71)=180 - 85 = 95$.

Step6: For Triangle G

Let the two unknown angles be $a$ and $b$. Since the sum of the interior angles of a triangle is 180° and we have one angle 121°, and assuming the triangle is isosceles (not given explicitly, but if we consider non - negative angles), if the triangle is isosceles, let $a=b$. Then $a=b=\frac{180 - 121}{2}=\frac{59}{2}=29.5$ (if non - isosceles, there are infinitely many solutions with the condition $a + b=180 - 121 = 59$ and $a>0,b>0$). For simplicity, if we assume equal non - given angles, the two angles are 29.5, 29.5.

Step7: For Triangle H

Let the two unknown angles be $m$ and $n$. We know $m + n=180 - 70=110$. If the triangle is isosceles, $m=n = 55$ (if non - isosceles, there are infinitely many solutions with $m>0,n>0$ and $m + n=110$). For simplicity, if we assume equal non - given angles, the two angles are 55, 55.

Answer:

Triangle C: 53; Triangle D: 37; Triangle E: 72; Triangle F: 95; Triangle G: 29.5, 29.5; Triangle H: 55, 55