Two triangles have equal bases. The altitude of one triangle is 3 units more than its base, and the altitude of the other is 3 units less than its base. Find the altitudes if the areas of the triangles differ by 21 square units.
Let the equal base = $b$. Altitude of Triangle 1: $h_1 = b + 3$. Altitude of Triangle 2: $h_2 = b - 3$.
A triangular piece of wood with dimensions 130 cm, 180 cm, and 190 cm is to be divided by a line bisecting the longest side drawn from its opposite vertex. Find the area of the part adjacent to the 180 cm side.
Step 1 — Total area (sides 130, 180, 190 cm; $s = 250$):
Find the area of the triangle whose sides are 12, 16, and 21 units.
Answer:
95.45 sq. units
87.45 sq. units
86.54 sq. units
82.78 sq. units
Use Heron's formula. The semiperimeter is: $s=\frac{12+16+21}{2}=24.5$ $A=\sqrt{s(s-12)(s-16)(s-21)}$ $A=\sqrt{24.5(12.5)(8.5)(3.5)}$ $\boxed{A=95.45\text{ sq. units}}$
If the sides of a parallelogram and an included angle are 6, 10, and 100º, respectively, find the length of the shorter diagonal.
Answer:
10.73
12.62
11.54
9.68
For sides 6 and 10 with included angle 100°, the diagonals satisfy: $d^2=a^2+b^2\pm2ab\cos100^\circ$ The shorter diagonal uses the plus sign because $\cos100^\circ$ is negative: $d^2=6^2+10^2+2(6)(10)\cos100^\circ$ $\boxed{d=10.73}$
Question Bank: t319
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
Determine which quadrant does the terminal side of the angle for the following conditions:
$\tan \theta > 0$ and $\sin \theta < 0$.
III quadrant
I quadrant
II quadrant
IV quadrant
$\cos \theta > 0$ and $\csc \theta < 0$.
I quadrant
II quadrant
IV quadrant
III quadrant
$\sec \theta < 0$ and $\cot \theta < 0$.
IV quadrant
I quadrant
III quadrant
II quadrant
Part 1.
$\sin\theta<0$ in Quadrants III and IV. $\tan\theta>0$ in Quadrants I and III. The common quadrant is III. $\boxed{\text{III quadrant}}$
Part 2.
$\cos\theta>0$ in Quadrants I and IV. $\csc\theta<0$ means $\sin\theta<0$, which occurs in Quadrants III and IV. The common quadrant is IV. $\boxed{\text{IV quadrant}}$
Part 3.
$\sec\theta<0$ means $\cos\theta<0$, in Quadrants II and III. $\cot\theta<0$ requires $\sin\theta$ and $\cos\theta$ to have opposite signs. With $\cos\theta<0$, this needs $\sin\theta>0$, i.e. Quadrant II. $\boxed{\text{II quadrant}}$
Question Bank: t349
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The interior angles of a triangular lot ABC are A = $80^\circ$ and B = $60^\circ$. Which of the following is correct? (Note: a, b, and c are the corresponding sides opposite angles A, B, and C, respectively)
b < a < c
c < b < a
c < b > a
a < c > b
The third angle is $C=180^\circ-80^\circ-60^\circ=40^\circ$. By the Law of Sines, larger angles face longer sides. Since $A(80^\circ)>B(60^\circ)>C(40^\circ)$, the sides satisfy $a>b>c$, i.e. $c$\boxed{c
Question Bank: t351
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
Given the sides of a triangle ABC: a = 36.3 cm, b = 23.9 cm, c = ?. The angle opposite side a is $102.7^\circ$. Compute the value of side c in centimeters.
15.25
22.57
25.78
20.45
By the Law of Sines, $\sin B=\dfrac{b\sin A}{a}=\dfrac{23.9\sin102.7^\circ}{36.3}=0.642$, so $B=39.97^\circ$. Then $C=180^\circ-102.7^\circ-39.97^\circ=37.33^\circ$. $c=\dfrac{a\sin C}{\sin A}=\dfrac{36.3\sin37.33^\circ}{\sin102.7^\circ}=22.57$ cm. $\boxed{22.57}$
Question Bank: t352
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The perimeter of a triangle is 271 cm. The interior angles measure $50^\circ$, $60^\circ$, and $70^\circ$, respectively. What is the length of the longest side of the triangle?
80.72 cm
91.26 cm
99.02 cm
93.65 cm
By the Law of Sines, sides are proportional to the sines of opposite angles. The longest side faces the largest angle, $70^\circ$. $\dfrac{c}{\text{perimeter}}=\dfrac{\sin70^\circ}{\sin50^\circ+\sin60^\circ+\sin70^\circ}=\dfrac{0.9397}{2.5717}$. $c=271\times\dfrac{0.9397}{2.5717}=99.02$ cm. $\boxed{99.02\text{ cm}}$
Question Bank: t353
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
Given the properties the sides and an angle of a triangle: a = 22 cm, c = 16 cm, C = $35^\circ$. How many distinct triangles can be formed?
1
2
0
undefined
SSA case. By the Law of Sines, $\sin A=\dfrac{a\sin C}{c}=\dfrac{22\sin35^\circ}{16}=0.789$. This gives $A=52.1^\circ$ or $A=127.9^\circ$. Both keep $A+C<180^\circ$ ($87.1^\circ$ and $162.9^\circ$), so both are valid. $\boxed{2}$
Question Bank: t354
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
Given the properties the sides and an angle of a triangle: a = 31 cm, b = 26 cm, C = $48^\circ$. How many distinct triangles can be formed?
1
2
0
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Here two sides ($a$, $b$) and the included angle $C$ are given. This is the SAS case. SAS always determines a unique triangle, so exactly one triangle can be formed. $\boxed{1}$
Question Bank: t355
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
A triangular lot ABC has the following data: Angle A = $40^\circ$, angle C = $60^\circ$, and side AC = 24 m. What is the length of side BC?
21.105 m
15.665 m
19.567 m
17.853 m
Side $AC$ is opposite $B$, where $B=180^\circ-40^\circ-60^\circ=80^\circ$. Side $BC$ is opposite $A=40^\circ$. By the Law of Sines, $BC=\dfrac{AC\sin A}{\sin B}=\dfrac{24\sin40^\circ}{\sin80^\circ}=15.665$ m. $\boxed{15.665\text{ m}}$
Question Bank: t356
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The length of the sides of a triangle are AB = 20 cm, BC = 24 cm, and AC = 17 cm. What is the angle opposite to side BC?
$44.31^\circ$
$83.54^\circ$
$55.26^\circ$
$80.43^\circ$
The angle opposite $BC=24$ is angle $A$, with the other two sides $b=AC=17$ and $c=AB=20$. Law of Cosines: $\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{17^2+20^2-24^2}{2(17)(20)}=\dfrac{113}{680}=0.1662$. $A=\cos^{-1}(0.1662)=80.43^\circ$. $\boxed{80.43^\circ}$
Question Bank: t358
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The second angle of a triangle is twice the first angle and the third angle is thrice the second angle. The perimeter of the triangle is 250 cm.
What is the value of the second angle of the triangle?
$20^\circ$
$60^\circ$
$40^\circ$
$120^\circ$
What is the length of the longest side of the triangle?
116.98 cm
86.82 cm
46.2 cm
132.76 cm
What is the length of the angle bisector to the longest side of the triangle?
32.32 cm
30.15 cm
28.54 cm
36.89 cm
Part 1.
Let the first angle be $x$. Then the second is $2x$ and the third is $3(2x)=6x$. $x+2x+6x=180^\circ\Rightarrow 9x=180^\circ\Rightarrow x=20^\circ$. The second angle is $2x=40^\circ$. $\boxed{40^\circ}$
Part 2.
The angles are $20^\circ$, $40^\circ$, and $120^\circ$. The longest side faces $120^\circ$. By the Law of Sines, sides are proportional to $\sin20^\circ,\sin40^\circ,\sin120^\circ$ summing to $1.8508$. Longest side $=250\times\dfrac{\sin120^\circ}{1.8508}=250\times\dfrac{0.8660}{1.8508}=116.98$ cm. $\boxed{116.98\text{ cm}}$
Part 3.
The longest side faces the $120^\circ$ vertex; the bisector is drawn from that vertex. The two sides forming it are $a=250\cdot\tfrac{\sin20^\circ}{1.8508}=46.20$ and $b=250\cdot\tfrac{\sin40^\circ}{1.8508}=86.83$ cm. Angle-bisector length: $t=\dfrac{2ab\cos(C/2)}{a+b}=\dfrac{2(46.20)(86.83)\cos60^\circ}{46.20+86.83}$. $t=\dfrac{46.20\times86.83}{133.03}=30.15$ cm. $\boxed{30.15\text{ cm}}$
Question Bank: t362
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
Two sides of a triangle measures 8 m and 12 m and the angle opposite the 8-m side is $30^\circ$. What is the measure of the third side of the triangle?
12.3 m
6.2 m
5.1 m
9.6 m
Let $a=8$ (opposite $A=30^\circ$) and $b=12$. By the Law of Sines, $\sin B=\dfrac{12\sin30^\circ}{8}=0.75$, so $B=48.6^\circ$ or $131.4^\circ$ (ambiguous case). Taking the obtuse case $B=131.4^\circ$: $C=180^\circ-30^\circ-131.4^\circ=18.6^\circ$. $c=\dfrac{a\sin C}{\sin A}=\dfrac{8\sin18.6^\circ}{\sin30^\circ}=5.1$ m (the value among the choices). $\boxed{5.1\text{ m}}$
Question Bank: t363
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The sides of a triangular lot are 28.45 m, 47.12 m, and 35.64 m. What is the angle opposite the 47.12 m side?
$93.98^\circ$
$98.51^\circ$
$82.34^\circ$
$87.67^\circ$
The angle opposite the longest side ($47.12$ m) is found by the Law of Cosines, with the other sides $28.45$ and $35.64$. $\cos\theta=\dfrac{28.45^2+35.64^2-47.12^2}{2(28.45)(35.64)}=\dfrac{-140.7}{2028.1}=-0.0694$. $\theta=\cos^{-1}(-0.0694)=93.98^\circ$. $\boxed{93.98^\circ}$
Question Bank: t364
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
In triangle ABC, side AB is 54 cm, side AC is 78 cm, and angle A is $56^\circ$.
What is the measure of side BC?
69.32 cm
65.49 cm
72.12 cm
57.89 cm
What is the measure of angle B?
$84.56^\circ$
$73.21^\circ$
$91.12^\circ$
$80.88^\circ$
What is the measure of angle C?
$43.12^\circ$
$49.43^\circ$
$38.67^\circ$
$51.21^\circ$
Part 1.
$BC$ is opposite angle $A$, with the two adjacent sides $AB=54$ and $AC=78$. Law of Cosines: $BC^2=54^2+78^2-2(54)(78)\cos56^\circ=9000-4711.5=4288.5$. $BC=\sqrt{4288.5}=65.49$ cm. $\boxed{65.49\text{ cm}}$
Part 2.
Angle $B$ is opposite $AC=78$. By the Law of Sines, $\sin B=\dfrac{78\sin56^\circ}{65.49}=0.987$. This gives $B=80.88^\circ$ (the supplement $99.12^\circ$ would make the angles exceed $180^\circ$). $\boxed{80.88^\circ}$
Part 3.
The angles of a triangle sum to $180^\circ$. $C=180^\circ-A-B=180^\circ-56^\circ-80.88^\circ=43.12^\circ$. $\boxed{43.12^\circ}$
Question Bank: t368
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
The three sides of a triangle measure 36 cm, 18 cm, and 24 cm. What is the length of the median drawn the longest side to opposite vertex.
12.325 cm
11.225 cm
10.125 cm
13.045 cm
The median to the longest side ($a=36$) uses $m_a=\dfrac{1}{2}\sqrt{2b^2+2c^2-a^2}$ with $b=18$, $c=24$. $m_a=\dfrac{1}{2}\sqrt{2(18^2)+2(24^2)-36^2}=\dfrac{1}{2}\sqrt{648+1152-1296}=\dfrac{1}{2}\sqrt{504}$. $m_a=11.225$ cm. $\boxed{11.225\text{ cm}}$
Question Bank: t369
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
If the perpendicular bisectors of two sides of a triangle meets at the third side, what kind of triangle is being described?
scalene triangle
obtuse triangle
right triangle
acute triangle
The perpendicular bisectors of the sides meet at the circumcenter. If this point lies on the third side, that side is a diameter of the circumscribed circle. By Thales' theorem, an angle inscribed in a semicircle is $90^\circ$, so the triangle is a right triangle. $\boxed{\text{right triangle}}$
Question Bank: t372
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
A flagpole 3 m high stands at the top of a pedestal 2 m high located at one side of a pathway. At the opposite side of the pathway directly facing the flagpole, the flagpole subtends the same angle as the pedestal. What is the width of the pathway?
4.47 m
3.21 m
6.28 m
8.41 m
Let $d$ be the path width. The pedestal (top at 2 m) subtends $\arctan\tfrac{2}{d}$; the flagpole (from 2 m to 5 m) subtends $\arctan\tfrac{5}{d}-\arctan\tfrac{2}{d}$. Setting these equal: $\arctan\tfrac{5}{d}=2\arctan\tfrac{2}{d}$. With $\tan\theta=\tfrac{2}{d}$, $\tan2\theta=\dfrac{4d}{d^2-4}=\dfrac{5}{d}$. $4d^2=5(d^2-4)\Rightarrow d^2=20\Rightarrow d=4.47$ m. $\boxed{4.47\text{ m}}$
Question Bank: t408
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
A spherical triangle has the following parts: $B = 42^\circ 30.2'$, $C = 128^\circ 46.6'$, $a = 64^\circ 26.4'$. Find the value of side $b$.
$57.24^\circ$
$63.25^\circ$
$46.47^\circ$
$103.99^\circ$
Side $a$ is the side between angles $B$ and $C$. Use the four-parts (cotangent) formula: $\cot b\,\sin a=\cos a\cos C+\sin C\cot B$. $\cot b=\dfrac{\cos64.44^\circ\cos128.78^\circ+\sin128.78^\circ\cot42.50^\circ}{\sin64.44^\circ}=\dfrac{-0.2705+0.8514}{0.9017}=0.644$. $b=\cot^{-1}(0.644)=57.24^\circ$. $\boxed{57.24^\circ}$
Question Bank: t409
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
In the spherical triangle ABC, $A = 116^\circ 19'$, $B = 55^\circ 30'$, and $C = 80^\circ 37'$. What is the value of side $b$?
$54^\circ 08'$
$56^\circ 02'$
$96^\circ 51'$
$47^\circ 08'$
With all three angles known, use the supplemental cosine rule for angles: $\cos B=-\cos A\cos C+\sin A\sin C\cos b$, so $\cos b=\dfrac{\cos B+\cos A\cos C}{\sin A\sin C}$. $\cos b=\dfrac{\cos55.5^\circ+\cos116.317^\circ\cos80.617^\circ}{\sin116.317^\circ\sin80.617^\circ}=\dfrac{0.5664-0.0724}{0.8843}=0.5587$. $b=\cos^{-1}(0.5587)=56.03^\circ=56^\circ02'$. $\boxed{56^\circ 02'}$
Question Bank: t410
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
In a spherical triangle, angle $B = 81^\circ 50'$ and angle $C = 94^\circ 30'$. If side $c = 90^\circ$, what is the value of angle $A$?
$56^\circ 28'$
$57^\circ 32'$
$62^\circ 24'$
$54^\circ 18'$
Use the supplemental cosine rule: $\cos C=-\cos A\cos B+\sin A\sin B\cos c$. Since $c=90^\circ$, $\cos c=0$, so $\cos C=-\cos A\cos B$, giving $\cos A=-\dfrac{\cos C}{\cos B}$. $\cos A=-\dfrac{\cos94.5^\circ}{\cos81.833^\circ}=-\dfrac{-0.0785}{0.1421}=0.5523$. $A=\cos^{-1}(0.5523)=56.47^\circ=56^\circ28'$. $\boxed{56^\circ 28'}$
Question Bank: t411
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
A spherical triangle have angles of $63^\circ$, $85^\circ$, and $54^\circ$. Find the area of the triangle if the sphere has a surface area of 16,286 square centimeters.
365.8 cm$^2$
287.4 cm$^2$
563.1 cm$^2$
497.6 cm$^2$
Spherical excess $E=(63^\circ+85^\circ+54^\circ)-180^\circ=22^\circ$. A spherical triangle's area is the fraction $\dfrac{E}{720^\circ}$ of the whole sphere: $\text{Area}=\dfrac{E}{720^\circ}\times(\text{surface area})$. $\text{Area}=\dfrac{22}{720}\times16{,}286=497.6$ cm$^2$. $\boxed{497.6\text{ cm}^2}$
Question Bank: t412
MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3
A triangle on the surface of a sphere has interior angles of $82^\circ$, $68^\circ$, and $105^\circ$. The volume of the sphere is 145,124.72 cc. What is the area of the triangle in square cm?
1874
1224
1563
1391
From the volume: $R=\sqrt[3]{\dfrac{3V}{4\pi}}=\sqrt[3]{\dfrac{3(145{,}124.72)}{4\pi}}=32.6$ cm, so surface area $=4\pi R^2=13{,}352$ cm$^2$. Spherical excess $E=(82^\circ+68^\circ+105^\circ)-180^\circ=75^\circ$. $\text{Area}=\dfrac{E}{720^\circ}\times4\pi R^2=\dfrac{75}{720}\times13{,}352=1391$ cm$^2$. $\boxed{1391}$
Question Bank: t2121
MSTE - Geometry and Trigonometry / Trigonometry / Besavilla CE Pre-Board Math & Surveying
The sides of a triangle are 45 m. and 55 m. long. If its area is 785.48 m$^2$, find the sum of the sides.
123 m.
135 m.
118 m.
132 m.
130 m.
Let the included angle between the 45 m and 55 m sides be $C$. $A=\frac{1}{2}ab\sin C$ $785.48=\frac{1}{2}(45)(55)\sin C$ $\sin C=\frac{2(785.48)}{45(55)}=0.6347$ Using the acute included angle gives the third side by the cosine law: $c^2=45^2+55^2-2(45)(55)\cos C$ $c\approx35$ m Sum of sides $=45+55+35$ $\boxed{135\text{ m}}$
Question Bank: t2125
MSTE - Geometry and Trigonometry / Trigonometry / Besavilla CE Pre-Board Math & Surveying
The perimeter of triangle ABC = 180 m. A = 46.567°, B = 104.478°, what is the dimension of the side opposite the biggest angle?
67 m.
63 m.
72 m.
75 m.
80 m.
The biggest angle is $B=104.478^\circ$, so the required side is $b$. First find $C$. $C=180^\circ-46.567^\circ-104.478^\circ=28.955^\circ$ By the sine rule, sides are proportional to the sines of their opposite angles, and the perimeter is 180 m. $b=180\frac{\sin104.478^\circ}{\sin46.567^\circ+\sin104.478^\circ+\sin28.955^\circ}$ $\boxed{b\approx80\text{ m}}$
Question Bank: w26
MSTE - Geometry and Trigonometry / Theorems on Triangles and Circles / MSTE May 2019
In triangle $ABC$, $AB = 30$ cm, $BC = 36$ cm, and $AC = 48$ cm. The perpendicular bisectors of the sides intersect at point $P$. How far is $P$ from side $BC$?
12.85 cm
15.92 cm
18.96 cm
22.54 cm
$P$ is the circumcenter; the circumradius is $r = \frac{abc}{4A_t}$. $s = \frac{36 + 48 + 30}{2} = 57$ $A_t = \sqrt{57(57-36)(57-48)(57-30)} = 539.32\text{ cm}^2$ $r = \frac{36 \cdot 48 \cdot 30}{4(539.32)} = 24.03\text{ cm}$ Distance from $P$ to $BC$ (half of $BC$ is 18): $x = \sqrt{24.03^2 - 18^2} = \boxed{15.92\text{ cm}}$
