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If the ratio of the three inner angles of a triangle is 9:4:5, then the triangle is ()
The inner angles of the triangle are:
180÷（9+4+5）×9＝10×9＝90°
180÷（9+4+5）×4＝10×4＝40°
180÷（9+4+5）×5＝10×5＝50°
Obviously, this triangle is a right triangle
The ratio of the degrees of the three inner angles of a triangle is 7:2:1 If two-thirds of number a is equal to number B, then number A: number b = ()
180/(7+2+1)=18
The three angles are:
18*7=126
18*2=36
18*1=18
This is a (obtuse) triangle
The degrees of the two inner angles of a triangle are 30 degrees and 46 degrees respectively. The degree of the third inner angle of a triangle is () and it is a triangle with () angle
The degrees of the two inner angles of a triangle are 30 degrees and 46 degrees respectively, and the degree of the third internal angle is (104 degrees). It is a (blunt) angle triangle
The ratio of the inner angles of a triangle is 7:2:1, and this triangle is () A. Obtuse triangle B. Acute triangle C. Right triangle
7+2+1=10，
180×7
10 = 126 degrees,
Because the maximum angle of the triangle is 126 degrees, the triangle is obtuse
Therefore, a
For a triangle, the degree ratio of the inner angles is 3:7:5. What are the inner angles of the triangle What triangle is this
Hello!
180÷（3+7+5）x3=36°
180÷（3+7+5）x7=84°
180÷（3+7+5）x5=60°
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The cosine of the base angle of an isosceles triangle is known to be 2 3, then the sine value of the vertex angle is () A. 2 Five Nine B. 4 Five Nine C. -4 Five Nine D. -2 Five Nine
If the base angle is α, then cos α = 2
3, the vertex angle is π - 2 α,
∴sinα=
1−cos2α=
Five
3，
∴sin（π-2α）=sin2α=2sinαcosα=2×
Five
3×2
3=4
Five
9．
Therefore, B is selected
The cosine of the vertex angle of an isosceles triangle is known to be 4 5, then the sine value of the base angle of the triangle is () A. Ten Ten B. − Ten Ten C. 3 Ten Ten D. −3 Ten Ten
If the base angle of the triangle is α, then the top angle is 180 ° - 2 α
∴cos（180°-2α）=-cos2α=4
Five
∴2sin2α-1=4
Five
∵ α is the inner angle of the triangle
∴sinα=3
Ten
Ten
Therefore, C
It is known that the sine of a base angle of an isosceles triangle is equal to 5 / 13 Urgent,
So, let ad = 5, ab = 13, and BD = 12, then we can know the area of ABC. Using the area formula, s = 1 / 2Ab * ac * (Sina), we can know the sinusoidal pull of the angle A. then we can just judge the other angle A
Δ ABC is an isosceles right triangle, ∠ a = 90; points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D
(1) It is proved that: the connection ad ∵ ABC is an isosceles right triangle, D is the midpoint of BC ∵ ad ⊥ BC, ad = BD = DC, ≌ DAQ = ∵ B and ∵ BP = AQ BPD ≌ ≌ △ aqd ∵ PD = QD, ≌ ADQ = ∠ BDP
As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° and points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D is the midpoint of BC (1) It is proved that △ PDQ is an isosceles right triangle; (2) When the point P moves to where, the quadrilateral apdq is square, and explain the reason
(1) It is proved that: connecting ad ∵ ABC is an isosceles right triangle, D is the midpoint of BC ∵ ad ⊥ BC, ad = BD = DC, ∵ DAQ = ∠ B. in △ BPD and △ aqd, BD = ad ∵ DBP ≌ ≌ △ aqd (SAS), PD = QD, ∵ ADQ = ≌ ≌≌△ aqd (SAS), PD = QD, ∠ ADQ = ∠ BDP, ? BDP + ∠ ADP = 90 °
RELATED INFORMATIONS
- 1. As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° and points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D is the midpoint of BC (1) It is proved that △ PDQ is an isosceles right triangle; (2) When the point P moves to where, the quadrilateral apdq is square, and explain the reason
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